10 - modular arithmetic

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    Art of Problem Solving

    WOOT

    Copyright AoPS Incorporated. This page is copyrighted material. You can view and print this page for your own use, but you

    cannot share the contents of this file with others.

    Class Transcript 01/21 - Modular Arithmetic B

    nsato 7:30:52 pm

    WOOT 2012-13: Modular Arithmetic B

    nsato 7:30:57 pm

    In today's lecture, we will continue our look at number theory and modular arithmetic. The first half of our class will focus on the use

    of Fermat's Little Theorem and Euler's Theorem. In the second half, we'll cover some more general number theory problems that

    are typical for this level.

    nsato 7:31:09 pm

    We'll start with a nice warm-up exercise.

    nsato 7:31:14 pm

    nsato 7:31:35 pm

    How can we find the missing digit?

    nsato 7:32:25 pm

    We can try to find the sum of the 9 digits. If all 10 digits were present, then their sum would be 0 + 1 + 2 + .. . + 9 = 45. However, we

    can't find the sum of the 9 digits without computing 2^29. Is there an easier way?

    delta1 7:33:00 pm

    mod 9

    Dunedubby 7:33:00 pm

    find the number mod 9

    Dunedubby 7:33:00 pm

    the number mod 9 is the same as the sum of the digits mod 9

    A123456789 7:33:00 pm

    Compute 2^29 (mod 9).

    aleph0 7:33:00 pmmod 9

    Binomial-theorem 7:33:00 pm

    Find the number modulo 9 to figure out which number we do not have in our sum

    brian22 7:33:02 pm

    mod 9!

    nsato 7:33:06 pm

    We know that a number and the sum of its digits are congruent modulo 9, so we can reduce 2^29 modulo 9. What do we get?

    AndroidFusion 7:34:08 pm

    5

    willwang123 7:34:08 pm

    5

    ProbaBillity 7:34:08 pm

    5

    zheng 7:34:08 pm

    5

    mjseaman1 7:34:14 pm

    5

    bestbearever 7:34:14 pm

    5

    Binomial-theorem 7:34:18 pm

    phi(9)=6, therefore 2^(29)=2^5=32=5(mod 9)

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    nsato 7:34:20 pm

    nsato 7:34:39 pm

    So what is the missing digit?

    mcdonalds106_7 7:35:25 pm

    4

    s.homberg 7:35:25 pm

    4

    math-fan 7:35:25 pm

    4

    distortedwalrus 7:35:25 pm

    4

    zheng 7:35:34 pm

    4

    PiCrazy31415 7:35:34 pm

    4

    mjseaman1 7:35:34 pm

    4

    MathisFun! 7:35:34 pm

    4

    Binomial-theorem 7:35:34 pm

    4!

    nsato 7:35:47 pm

    The missing digit must be congruent to -5 modulo 9, so the missing digit is 4.

    ProbaBillity 7:36:01 pm

    and indeed it is 4; 2^29 = 536870912.

    nsato 7:36:03 pm

    Indeed, 2^29 = 536870912. All digits are present except 4.

    nsato 7:36:16 pm

    nsato 7:36:28 pm

    How do we start?

    ProbaBillity 7:37:01 pm

    mod 1000

    trophies 7:37:01 pm

    reduce it modulo 1000

    bobcat120 7:37:01 pm

    take it mod 1000

    PiCrazy31415 7:37:01 pm

    modulo 1000

    bestbearever 7:37:01 pm

    mod 1000

    nsato 7:37:04 pm

    Finding the last three digits of a number is an indication to work modulo 1000.

    nsato 7:37:10 pm

    nsato 7:37:14 pm

    How can we reduce this?

    pgmath 7:37:53 pm

    Euler's Theorem

    ProbaBillity 7:37:53 pm

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    Euler's theorem

    nsato 7:38:15 pm

    We can use Euler's Theorem.

    trophies 7:38:20 pm

    9^400= 1(mod 1000) by euler's theorem

    A123456789 7:38:20 pm

    9^400=1(mod 100), so compute 8^7(mod 400).

    distortedwalrus 7:38:20 pm

    note that 9^400 == 1 (mod 1000)

    nsato 7:38:30 pm

    mentalgenius 7:38:44 pm

    well phi(1000) = 400, so we want 8^7 mod 400

    Dunedubby 7:38:44 pm

    find 8^7 mod phi(1000)

    nsato 7:38:54 pm

    nsato 7:39:02 pm

    How can we do this?

    diger 7:40:20 pm

    eulers theorem again

    distortedwalrus 7:40:20 pm

    use euler's theorem again

    nsato 7:40:22 pm

    Euler's Theorem doesn't help, because the exponent 7 is far less than phi(400).

    brian22 7:40:31 pm

    8^3*8^3*8

    trophies 7:40:31 pm

    Note that 8^3=512= 112(mod 400). Thus, we have 112*112*8= 112*96= 352(mod 1000)

    nsato 7:40:44 pm

    Probably the easiest way to compute this is to look at the first few powers of 8 modulo 400. We do this by starting with 1, then

    multiplying each term by 8 and reducing modulo 400.

    nsato 7:41:05 pm

    This gives us 1, 8, 64, 112, 96, 368, 144, 352.

    nsato 7:41:16 pm

    (Or we can combine powers, as suggested above.)

    mentalgenius 7:41:28 pm

    8^3 = 512 112 (mod 400)

    nsato 7:41:37 pm

    nsato 7:41:56 pm

    How can we compute this residue?

    nsato 7:42:29 pm

    We have gone as far as we can with Euler's Theorem. We can try to compute this by looking at the first few powers of 9 modulo

    1000, but there is an easier way.

    nsato 7:42:50 pm

    Note that 9 is close to 10, and 1000 = 10^3. Can we exploit this fact somehow?

    trophies 7:43:28 pm

    binomial theorem!

    TheLittleOne 7:43:28 pm

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    nsato 7:47:03 pm

    We use the prime factorization of 30 = 2 x 3 x 5. Hence, it suffices to show that n^5 - n is divisible by 2, 3, and 5, for all integers n.

    nsato 7:47:22 pm

    Is n 5 - n divisible by 2?

    ProbaBillity 7:48:20 pm

    yes, we test both 0 and 1 and they work mod 2.

    bobcat120 7:48:20 pm

    yes, n^5 and n are the same parity

    pgmath 7:48:20 pm

    Yes, n^5 and n have the same parity

    trophies 7:48:20 pm

    Yes... we can check 1(mod 2) and 0(mod 2), which both work.

    zheng 7:48:20 pm

    Yes. If n is odd we get odd-odd=even. If n is even we get even-even=even

    Showpar 7:48:20 pm

    Yes, n^5 and n have the same modulo 2 residue

    mentalgenius 7:48:20 pm

    two cases - n even or n odd; if n even, trivial; if n odd, 1-1 0 (mod 2), so yes

    aleph0 7:48:20 pm

    n^5 and n must have same parity

    nsato 7:48:23 pm

    Yes. By reducing modulo 2, we can check for n = 0 and n = 1.

    nsato 7:48:26 pm

    Is n 5 - n divisible by 3?

    ProbaBillity 7:49:20 pm

    0, 1, and 2 work. So it is.

    trophies 7:49:20 pm

    Yes, we can check 0(mod 3), 1(mod 3), and 2(mod 3)

    mjseaman1 7:49:20 pm

    Yes check n==0,1,2 mod 3

    distortedwalrus 7:49:20 pmyes by checking 0, 1, and 2

    binmu 7:49:20 pm

    yes, check 0, 1 and 2(mod3)

    nsato 7:49:24 pm

    Yes. By reducing modulo 3, we can check for n = 0, 1, and 2.

    nsato 7:49:27 pm

    Is n 5 - n divisible by 5?

    ProbaBillity 7:50:15 pm

    0, 1, 2, 3, and 4 work. Thus 5|n^5 - n. And we are done!

    pgmath 7:50:15 pm

    Yes, by FLT

    zheng 7:50:15 pm

    Fermat's little gives n-n so yes

    PiCrazy31415 7:50:15 pm

    yes, fermat's little theorem

    brian22 7:50:15 pm

    by Fermat's litt le theorem

    Binomial-theorem 7:50:15 pm

    mjseaman1 7:50:15 pm

    Yes by Fermat

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    trophies 7:50:15 pm

    Yes, we can use Fermat's little theorem... n^5=n(mod 5)-n(mod 5)= 0(mod 5)

    diger 7:50:15 pm

    fermats little theorem

    nsato 7:50:18 pm

    Yes. By reducing modulo 5, we can check for n = 0, 1, 2, 3, and 4. We can also simply cite Fermat's Little Theorem for the prime p =

    5.

    nsato 7:51:04 pm

    Hence, n^5 - n is divisible by 2 x 3 x 5 = 30 for all integers n.

    nsato 7:51:09 pm

    This simple problem illustrates an important principle. If you want to prove that a number is divisible by n, it helps to look at the

    individual prime powers of n.

    nsato 7:51:33 pm

    nsato 7:52:25 pm

    We can start by trying to reduce the given expression modulo 2000, but all the bases are already less than 2000. So what else can

    we do?

    ProbaBillity 7:53:37 pm

    divisible by 16 and 125

    zheng 7:53:37 pm

    use mod 16 and mod 125

    brian22 7:53:37 pm

    mod 16 and 125

    pgmath 7:53:37 pm

    modulo 2^4 = 16 and 5^3 = 125

    mjseaman1 7:53:37 pm

    Reduce it mod 16 and 125. If it is zero mod both, it is divisible by 2000.

    distortedwalrus 7:53:37 pm

    note that 2000=2^4*5^3

    Binomial-theorem 7:53:37 pm

    2000=2*1000=2^4*5^3 look mod 16 and mod 125

    nsato 7:53:49 pm

    The prime factorization of 2000 is 2 4 x 5^3, so we look at the factors 2 4 = 16 and 5^3 = 125 separately.

    nsato 7:54:00 pm

    How does the given expression reduce modulo 16?

    PiCrazy31415 7:55:21 pm

    0 modulo 16

    diger 7:55:21 pm

    9^n - 9^n +12^n -(-4)^n == 0

    Binomial-theorem 7:55:21 pm9^n-9^n+(-4)^n-(-4)^n=0(mod 16) YAY!

    brian22 7:55:21 pm

    0

    zheng 7:55:21 pm

    9^n-9^n+12^n-12^n=0

    distortedwalrus 7:55:21 pm

    it's 0 (mod 16)

    tc1729 7:55:25 pm

    9^n - 9^n + 12^n - 12^n = 0

    nsato 7:55:31 pm

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    nsato 7:55:40 pm

    Therefore, 121^n - 25^n + 1900 n - (-4) n is divisible by 16.

    nsato 7:55:51 pm

    How does the given expression reduce modulo 125?

    mentalgenius 7:56:59 pm

    0

    pgmath 7:56:59 pm

    bestbearever 7:56:59 pm

    0 mod 125 (cancels)

    Binomial-theorem 7:56:59 pm

    (-4)^n-25^n+25^n-(-4)^n so again 0

    PiCrazy31415 7:56:59 pm

    0 modulo 125

    tc1729 7:56:59 pm

    121^n - 25^n + 25^n -121^n = 0 (mod 125)

    trophies 7:56:59 pm

    121^n-25^n+25^n-121^n= 0(mod 125).

    zheng 7:56:59 pm

    (-4)^n-25^n+25^n-(-4)^n=0

    A123456789 7:56:59 pm

    (-4)^n-25^n+25^n-(-4)^n

    Showpar 7:56:59 pm

    (-4)^n-25^n+25^n-(-4)^n

    ProbaBillity 7:57:03 pm

    (-4^n) - 25^n + 25^n - (-4^n) 0 (mod 25) DOUBLE YAY! WE DONE! that was really nice.

    nsato 7:57:07 pm

    nsato 7:57:18 pm

    Therefore, 121^n - 25^n + 1900 n - (-4) n is divisible by 125.

    nsato 7:57:27 pm

    We conclude that 121 n - 25^n + 1900 n - (-4)^n is divisible by 2000 for all positive integers n.

    nsato 7:57:51 pm

    Again, all we had to do was look at individual prime powers, which helps break down the expression.

    nsato 7:58:02 pm

    nsato 7:58:14 pm

    What can we do with this equation?

    nsato 7:59:22 pm

    All those fifth powers... does that remind us anything?

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    Dunedubby 7:59:52 pm

    take it mod 30!

    ProbaBillity 7:59:52 pm

    n^5 n (mod 30)

    Dunedubby 7:59:52 pm

    we know that n^5 == n mod 30

    nsato 8:00:12 pm

    We've shown that n^5 is congruent to n modulo 30, so we can reduce the given equation modulo 30. What does that tell us about n

    ProbaBillity 8:01:57 pm

    n 24 (mod 30)

    mentalgenius 8:01:57 pm

    n 24 (mod 30)

    zheng 8:01:57 pm

    n=133+110+84+27=24 (mod 30)

    steve314 8:01:57 pm

    n==24 mod 30

    brian22 8:01:57 pm

    13+20+24+27==22==n mod 30

    nsato 8:02:09 pm

    nsato 8:03:04 pm

    That helps narrow down the possible values of n. What else can we do to help find n?

    ged3.14 8:03:56 pm

    use units digits along with bounding

    Showpar 8:03:56 pm

    Set bounds on n.

    ged3.14 8:04:01 pm

    then bound n

    Dunedubby 8:04:01 pm

    set an upper bound and lower bound

    nsato 8:04:02 pm

    Let's try to find some bounds on n. What's one obvious bound for n?

    PiCrazy31415 8:04:57 pm

    larger than 133

    zheng 8:04:57 pm

    n>133

    steve314 8:04:57 pm

    n>133

    Showpar 8:04:57 pmn>133

    flappingwings 8:04:57 pm

    n is definitely greater than 133

    diger 8:04:57 pm

    n>133

    bestbearever 8:04:57 pm

    greater than 133

    Binomial-theorem 8:04:57 pm

    n>133

    yangdongyan 8:04:57 pm

    n>133

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    nsato 8:04:59 pm

    Clearly n > 133.

    nsato 8:05:11 pm

    Now we need to find an upper bound on n.

    nsato 8:05:32 pm

    The largest term in the left-hand side is 133^5, so we compare all the other terms to 133^5.

    nsato 8:05:42 pm

    We know that 110^5 < 133^5.

    nsato 8:05:53 pm

    Can we compare 84^5 + 27 5 and 133^5?

    PiCrazy31415 8:07:17 pm

    84^5 + 27^5 < 133^5

    flappingwings 8:07:17 pm

    since 84+27 is less than 133, the former is less than the latter

    Dunedubby 8:07:17 pm

    84^5 + 27^5 < (84 + 27)^5 < 133^5

    King6997 8:07:17 pm

    84^5+27^5 < 133^5

    tc1729 8:07:17 pm

    distortedwalrus 8:07:17 pm

    well it's less than (84+27)^5=111^5

    Showpar 8:07:19 pm

    84^5+27^5

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    5^5 and 4^5

    ProbaBillity 8:12:48 pm

    3125/1024

    steve314 8:12:51 pm

    5^5/4^5

    distortedwalrus 8:12:51 pm

    yeah 5^5/4^5

    tc1729 8:12:55 pm

    5^5/4^5

    nsato 8:13:01 pm

    We see that 5^5/4 5 = 3125/1024 is approximately 3.

    nsato 8:13:14 pm

    Even better, this ratio is slightly larger than 3, which means we can use this ratio to get an upper bound on n:

    nsato 8:13:27 pm

    nsato 8:13:41 pm

    Therefore, n < 5/4 x 133 = 166 + 1/4.

    nsato 8:14:06 pm

    We also know that n > 133 and n is congruent to 24 modulo 30. Do we have enough information to determine n?

    trophies 8:14:58 pm

    thus, n must be 144

    bestbearever 8:14:58 pm

    144

    AndroidFusion 8:14:58 pm

    n = 144

    mjseaman1 8:14:58 pm

    Yes n is 144

    Binomial-theorem 8:14:58 pm

    n=144, 174, ... but 174>166 so n=144

    mentalgenius 8:14:58 pm

    yes, n = 144

    A123456789 8:14:58 pm

    n=144

    nsato 8:15:01 pm

    The only n satisfying 133 < n < 166 + 1/4, and n congruent to 24 modulo 30 is n = 144.

    nsato 8:15:39 pm

    nsato 8:16:03 pm

    What makes this sum difficult to deal with?

    ProbaBillity 8:17:17 pm

    the absolute values

    delta1 8:17:17 pm

    absolute value

    diger 8:17:17 pm

    absolute values

    math-fan 8:17:17 pm

    all absolute values

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    trophies 8:27:27 pm

    k=1(mod 2), or in other words, k is odd.

    steve314 8:27:27 pm

    k is odd

    Showpar 8:27:27 pm

    LHS must be even, so k is odd

    mjseaman1 8:27:27 pm

    k is odd

    nsato 8:27:31 pm

    The left-hand side must be even. Therefore, 5k must be odd, which means k must be odd.

    nsato 8:27:44 pm

    Let k = 2n + 1 for some integer n.

    nsato 8:27:47 pm

    Now what?

    bestbearever 8:28:55 pm

    combine that with the very first 5994-5k

    binmu 8:28:55 pm

    plug back in

    AndroidFusion 8:28:55 pm

    so 5994 - 5k = 5989 - 10k is congruent to 9 mod 10

    ProbaBillity 8:28:55 pm

    plug it in

    willwang123 8:28:55 pm

    substitute that in

    binmu 8:28:55 pm

    5994-5(2n+1)=5994-10n+5==9(mod10)

    Dunedubby 8:29:00 pm

    so we're done! - plugging in the first equation 5994 - 5k means that it must end in a 9

    nsato 8:29:02 pm

    We can substitute into the equation above, to get

    nsato 8:29:06 pm

    nsato 8:29:28 pm

    It looks like we can conclude that this number always ends in 9, but there is a small catch. What's the catch?

    ProbaBillity 8:30:26 pm

    we must verify that n is less than 599

    A123456789 8:30:26 pm

    It could be negative.

    willwang123 8:30:26 pm

    or if n>=599

    Calebhe1290 8:30:26 pm

    5989

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    mjseaman1 8:31:45 pm

    The sum is positive because it the sum of absolute values each of which is 1 or 6.

    s.homberg 8:31:45 pm

    each b_i > a_i, so the sum must be positive

    Dunedubby 8:31:51 pm

    absolute value

    nsato 8:31:58 pm

    The original sum was expressed as a sum of absolute values, which must be nonnegative. And a nonnegative number of the form

    5989 - 10n must end in 9, so we are home free.

    ProbaBillity 8:32:07 pm

    But it is, since 2n + 1 = k

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    bobcat120 8:42:14 pm

    x must be odd

    AndroidFusion 8:42:14 pm

    x is odd

    nsato 8:42:17 pm

    The number 2d - 1 is always odd, so x is odd.

    nsato 8:42:36 pm

    Since x is odd, what can we do?

    Calebhe1290 8:43:27 pm

    write it as 2n+1

    PiCrazy31415 8:43:27 pm

    let x = 2n+1

    Dunedubby 8:43:27 pm

    express as 2n + 1

    willwang123 8:43:27 pm

    say it's 2n+1

    binmu 8:43:27 pm

    x=2n+1

    A123456789 8:43:27 pm

    x=2n+1

    AndroidFusion 8:43:27 pm

    substitute it for 2n+1

    nsato 8:43:30 pm

    Whenever we find some fact like this, we express it algebraically.

    nsato 8:43:36 pm

    Let x = 2n + 1 for some integer n. Then 2d - 1 = x^2 = (2n + 1)^2 = 4n 2 + 4n + 1.

    zheng 8:44:41 pm

    d=2n^2+2n+1

    ProbaBillity 8:44:41 pm

    so d = 2n^2 + 2n + 1. Time to plug more things in!

    Binomial-theorem 8:44:41 pmd=2n^2+2n+1

    soy_un_chemisto 8:44:41 pm

    then d is 2n^2 + 2n + 1

    Dunedubby 8:44:41 pm

    therefore d = 2n^2 + 2n + 1

    nsato 8:44:46 pm

    Then d = 2n 2 + 2n + 1. What does this tell us about d?

    esque 8:45:42 pm

    d is odd

    distortedwalrus 8:45:42 pm

    d is odd

    soy_un_chemisto 8:45:42 pm

    d is odd

    Calebhe1290 8:45:42 pm

    d is odd

    zheng 8:45:42 pm

    d is odd

    nsato 8:45:51 pm

    This equation says that d is odd.

    nsato 8:45:57 pm

    Then what can we say about 5d - 1 = y^2 and 13d - 1 = z^2?

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    pgmath 8:46:50 pm

    5d -1 and 13d - 1 are both even

    delta1 8:46:58 pm

    y and z are even

    Binomial-theorem 8:46:58 pm

    y and z are even

    Showpar 8:46:58 pm

    y and z are both even

    mjseaman1 8:46:58 pm

    y and z are even

    bestbearever 8:46:58 pm

    y and z are even

    mentalgenius 8:46:58 pm

    y and z are both even

    nsato 8:47:04 pm

    Since d is odd, 5d - 1 = y^2 and 13d - 1 = z^2 are even. Hence, y and z are even.

    nsato 8:47:25 pm

    Again, let's express these statements algebraically, by introducing variables.

    brian22 8:47:44 pm

    y=2k, z=2j

    nsato 8:47:52 pm

    Let y = 2u and z = 2v. Then 5d - 1 = y^2 = 4u 2 and 13d - 1 = 4v^2.

    nsato 8:48:03 pm

    What can we do with these equations?

    delta1 8:49:01 pm

    subtract them

    A123456789 8:49:01 pm

    Subtract the first from the second.

    diger 8:49:01 pm

    subtract

    binmu 8:49:01 pmsubtract

    Calebhe1290 8:49:01 pm

    subtract

    ProbaBillity 8:49:01 pm

    subtract them

    pgmath 8:49:06 pm

    subtract the first from the second

    nsato 8:49:07 pm

    We can subtract them, to get 8d = 4v^2 - 4u^2, so 2d = v^2 - u^2.

    nsato 8:49:18 pm

    What can we do with this equation?

    brian22 8:50:15 pm

    factor!

    willwang123 8:50:15 pm

    diff of squares

    binmu 8:50:15 pm

    2d=(v+u)(v-u)

    Calebhe1290 8:50:15 pm

    factor the rhs

    PiCrazy31415 8:50:15 pm

    factorize

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    soy_un_chemisto 8:50:17 pm

    factor the right side

    nsato 8:50:19 pm

    We can factor the right-hand side, to get 2d = (v + u)(v - u).

    nsato 8:50:25 pm

    Can we say anything interesting about the factors v + u and v - u?

    steve314 8:51:19 pm

    they have the same parity

    Dunedubby 8:51:19 pm

    they are of the same parity

    diger 8:51:19 pm

    same parity

    vjnmath 8:51:19 pm

    They must be the same parity

    Binomial-theorem 8:51:19 pm

    same parity

    pgmath 8:51:19 pm

    they have the same parity

    ProbaBillity 8:51:24 pm

    same parity

    brian22 8:51:24 pm

    same parity

    nsato 8:51:28 pm

    Either v + u and v - u are both even or both odd (since their difference 2u is an even number).

    nsato 8:51:55 pm

    What happens if both v + u and v - u are even?

    binmu 8:52:47 pm

    2d is divisible by 4

    bestbearever 8:52:47 pm

    2d is congruent to 0 mod 4

    zheng 8:52:47 pmthen we have that d must be even, contradiction

    distortedwalrus 8:52:47 pm

    both sides are multiples of 4, so d is even

    PiCrazy31415 8:52:47 pm

    then 2d is divisible by 4, but d is odd

    diger 8:52:47 pm

    2d is a multiple of 4 and d is even

    willwang123 8:52:47 pm

    d is even too

    cire_il 8:52:47 pm

    then it means d is even

    Binomial-theorem 8:52:47 pm

    then d is even, but that's a contradiction

    Dunedubby 8:52:47 pm

    then d is even, a contradiction

    nsato 8:52:53 pm

    If both v + u and v - u are even, then 2d = (v + u)(v - u) is a multiple of 4, which is impossible because d is odd.

    nsato 8:53:03 pm

    What happens if both v + u and v - u are odd?

    ProbaBillity 8:53:56 pm

    another contradict ion. LHS is 2d which is even, RHS is odd.

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    diger 8:53:56 pm

    2d is odd

    bestbearever 8:53:56 pm

    2d must be odd

    PiCrazy31415 8:53:56 pm

    then 2d is odd, contradiction

    binmu 8:53:56 pm

    2d is odd too

    Binomial-theorem 8:53:56 pm

    then 2d must be odd, absurd again

    tc1729 8:53:56 pm

    d is odd

    Showpar 8:53:56 pm

    The RHS is odd, while the LHS is even, contradiction

    nsato 8:54:00 pm

    If both v + u and v - u are odd, then 2d = (v + u)(v - u) is odd, which is again impossible because 2d is even.

    nsato 8:54:22 pm

    We have obtained a contradiction, so at least one of 2d - 1, 5d - 1, and 13d - 1 is not a perfect square.

    nsato 8:54:41 pm

    It can seem simple, but using parity can be a powerful tool. And whenever you find that a number is even (odd resp.), it is usually a

    good idea to write it in the form 2n (2n + 1 resp.).

    nsato 8:54:54 pm

    nsato 8:55:19 pm

    You may recall Euclid's proof that there are an infinite number of primes. How can we start?

    Binomial-theorem 8:55:58 pm

    contradiction again!

    tc1729 8:55:58 pm

    assume there are only finitely many

    PiCrazy31415 8:55:58 pm

    proof by contradiction

    A123456789 8:55:58 pm

    Contradiction.

    bestbearever 8:55:58 pm

    assume there are a finite number

    distortedwalrus 8:55:58 pm

    we can use a contradiction again

    nsato 8:56:03 pm

    We argue by contradiction.

    zheng 8:56:08 pm

    assume there are finite primes of the form 3n+2

    nsato 8:56:10 pm

    nsato 8:56:23 pm

    Then what?

    willwang123 8:57:00 pm

    multiply them

    mentalgenius 8:57:00 pm

    multiply them all together and see what you get

    nsato 8:57:05 pm

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    Is there any way to fix this argument?

    Binomial-theorem 9:04:31 pm

    Consider 3p_1p_2p_3...p_k-1

    nsato 9:04:36 pm

    tc1729 9:05:07 pm

    then it has to be 2 mod 3

    nsato 9:05:10 pm

    Then P always reduces to 2 modulo 3, and it is relatively prime to all the p_i, so the argument that we used above now works.

    nsato 9:05:28 pm

    More generally, if a and b are relatively prime positive integers, then there are infinitely many primes of the form an + b. This result

    is known as Dirichlet Theorem, but it is an advanced result and is difficult to prove.

    nsato 9:06:24 pm

    The idea of Euclid's proof is also useful for the following problem.

    nsato 9:06:30 pm

    Binomial-theorem 9:07:11 pm

    that looks complex

    nsato 9:07:12 pm

    It looks hard; we just have to take things one step at a time.

    nsato 9:08:19 pm

    Let's start by looking at small primes. Is the prime 2 in M?

    nsato 9:08:51 pm

    Since M contains at least three primes, there is an odd prime in M, say p. Let A = {p}. Then by the condition in the problem, all the

    prime factors of p - 1 are also in M.

    binmu 9:09:46 pm

    so yes 2 is in M

    zheng 9:09:46 pm

    Since p is odd, p-1 is even so 2 must be in M

    Calebhe1290 9:09:46 pm

    p-1 is even, so 2 must be in M

    mentalgenius 9:09:52 pm

    so 2 has to be in m

    nsato 9:09:56 pm

    But p - 1 is even, so 2 is in M.

    nsato 9:10:22 pm

    Can we show that 3 must be in M? Would cases would easily lead to 3 being in M?

    Dunedubby 9:11:21 pm

    3n + 1 prime

    math-fan 9:11:21 pm

    primes that are equal to 1 mod 3

    Binomial-theorem 9:11:21 pm

    p=1(mod 3)

    nsato 9:11:28 pm

    If M contains a prime of the form p = 3n + 1, then take A = {p}. Then all the factors of p - 1 = 3n are also in M, so 3 is in M.

    nsato 9:12:46 pm

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    Otherwise, M contains two primes of the form 3n + 2, say p and q. What should we take the set A as in this case?

    Dunedubby 9:13:20 pm

    and if not, then the product of two 3n + 2s

    capu 9:13:33 pm

    If there is p congruent to 1 mod 3 then done (A= that prime). If not then there are two primes congruent to 2 mod 3 then A= those two and

    done. Therefore 3 is in M

    zheng 9:13:33 pm

    {2, p} or {2,q}

    diger 9:13:33 pm

    A = {p,q}

    ProbaBillity 9:13:33 pm

    {p, q}

    nsato 9:13:40 pm

    We can take A = {p, q}. Then all the prime factors of pq - 1 are also in M.

    mentalgenius 9:13:55 pm

    pq 1 (mod 3), so 3 is guaranteed to be in M

    nsato 9:13:56 pm

    But pq - 1 is congruent to 2 x 2 - 1 = 3, or 0 modulo 3, so pq - 1 is divisible by 3. Therefore, 3 is in M. In either case, 3 is in M.

    nsato 9:14:39 pm

    So now we know that 2 and 3 must be in M, which is a great start. What does this tell us?

    Dunedubby 9:15:42 pm

    and {2, 3} so 5 is in M as well

    distortedwalrus 9:15:42 pm

    5=2*3-1 is in M

    diger 9:15:42 pm

    5 must be in M for when A = {2,3}

    Binomial-theorem 9:15:42 pm

    5 is in M, becuase -1+2*3=5

    zheng 9:15:42 pm

    2*3-1=5 so 5 must be in M

    nsato 9:15:44 pm

    Taking A = {2,3}, we see that 5 must be in M.

    nsato 9:15:56 pm

    So now we have 2, 3, and 5. Anything else?

    ProbaBillity 9:16:36 pm

    7 is in M, because 3*5 - 1 = 14 which has 7 as a prime factor

    Dunedubby 9:16:36 pm

    and {3, 5} so 7

    zheng 9:16:36 pm

    A={3,5} gives that 7 must be in M

    diger 9:16:36 pm

    if A = {3,5} we get 7

    soy_un_chemisto 9:16:36 pm

    3 * 5 - 1 = 14 so 7 must also be in M

    binmu 9:16:36 pm

    7 is in M, 5*3-1=14

    steve314 9:16:36 pm

    3*5-1 =14=2*7 so 7 is also in m

    Calebhe1290 9:16:36 pm

    7, {3,5}

    yankeesfan 9:16:36 pm

    3*5-1=14 so 7

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    nsato 9:16:39 pm

    Taking A = {3,5}, we see that 7 must be in M.

    nsato 9:17:01 pm

    Hence, the first few primes are in M. We want to prove that M contains all primes.

    Binomial-theorem 9:17:11 pm

    we can follow this logic, can't we?

    nsato 9:17:40 pm

    We can generate more primes, but we'll need a general approach. Obviously, we can't do this every prime!

    nsato 9:17:55 pm

    But primes are hard to get a hold of.

    nsato 9:17:58 pm

    What intermediate result can we try proving first?

    nsato 9:18:15 pm

    (Here, you might think of Euclid's proof...)

    PiCrazy31415 9:18:46 pm

    M is infinite?

    binmu 9:18:46 pm

    infinite number of primes in M

    mentalgenius 9:18:46 pm

    there are an infinite number of primes in M

    Dunedubby 9:18:46 pm

    there an infinite number of primes in M

    diger 9:18:46 pm

    M has infinite elements

    aleph0 9:18:46 pm

    M is infinite

    nsato 9:18:54 pm

    Let's try proving first that M is infinite.

    nsato 9:19:03 pm

    nsato 9:19:26 pm

    As a start, let's take A to be all the primes in M, except one, say p_i. (Remember that A must be a proper subset of M.)

    nsato 9:19:42 pm

    nsato 9:20:01 pm

    pgmath 9:20:58 pm

    It is an integer power of p_i

    diger 9:20:58 pm

    it is a power of p_i

    aleph0 9:20:58 pm

    it's a power of p_i

    delta1 9:20:58 pm

    it is some power of p_i

    nsato 9:21:06 pm

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    nsato 9:24:22 pm

    Let's start analyzing this equation.

    nsato 9:24:38 pm

    nsato 9:24:47 pm

    Also, P is at least 2 x 3 x 5 x 7. So what can we say about a?

    mjseaman1 9:27:35 pm

    PiCrazy31415 9:27:35 pm

    a7

    Calebhe1290 9:27:35 pm

    a is at least 7

    ProbaBillity 9:27:35 pm

    a is at least 7

    nsato 9:28:05 pm

    nsato 9:28:10 pm

    nsato 9:28:26 pm

    So what can we say about b?

    brian22 9:29:14 pm

    b=1

    mjseaman1 9:29:14 pm

    b=1

    delta1 9:29:14 pm

    it is 1

    capu 9:29:14 pm

    b=1

    PiCrazy31415 9:29:14 pm

    b=1

    diger 9:29:14 pm

    divide by 2 to see b=1

    nsato 9:29:23 pm

    Right, because 2^a - 2 has exactly one factor of 2.

    nsato 9:29:35 pm

    nsato 9:29:41 pm

    How can we analyze this equation?

    PiCrazy31415 9:30:28 pm

    modulo 3

    AndroidFusion 9:30:28 pm

    mod 3

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    nsato 9:30:29 pm

    We can try reducing modulo 3.

    nsato 9:30:35 pm

    nsato 9:31:03 pm

    What does this congruence tell us about a?

    soy_un_chemisto 9:31:30 pm

    so a is odd

    mentalgenius 9:31:30 pm

    a is odd

    A123456789 9:31:30 pm

    a is odd.

    distortedwalrus 9:31:30 pm

    a is odd

    Showpar 9:31:30 pm

    a is odd

    trophies 9:31:30 pm

    a is odd

    cire_il 9:31:30 pm

    a is odd

    ProbaBillity 9:31:34 pm

    a = 2n+1

    nsato 9:31:38 pm

    This congruence tells us that a - 1 is even. Let a - 1 = 2n.

    nsato 9:31:48 pm

    nsato 9:31:55 pm

    What can we do with this equation?

    distortedwalrus 9:32:44 pm

    factor as difference of sqaures

    binmu 9:32:44 pm

    (2^n-1)(2^n+1)

    A123456789 9:32:44 pm

    Factor the right side of the equation.

    nsato 9:32:47 pm

    nsato 9:32:52 pm

    What does this equation say?

    delta1 9:33:41 pm

    the two factors are both powers of 3

    mjseaman1 9:33:41 pm

    Both of the factors on the RHS are powers of 3

    nsato 9:33:52 pm

    This equation says that both 2^n + 1 and 2^n - 1 are powers of 3.

    nsato 9:33:57 pm

    But (2^n + 1) - (2^n - 1) = 2.

    Calebhe1290 9:34:20 pm

    only 3 and 1 would work

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    Dunedubby 9:34:31 pm

    that 2^n + 1 and 2^n - 1 are both powers of 3, so n = 1

    A123456789 9:34:31 pm

    n=1.

    soy_un_chemisto 9:34:31 pm

    2^n + 1 and 2^n - 1 must be powers of 3, but this is only satisfied when n = 1

    delta1 9:34:31 pm

    so n=1

    ProbaBillity 9:34:31 pm

    n = 1

    nsato 9:34:40 pm

    The only powers of 3 that differ by 2 are 3 and 1, so n = 1, and a = 3.

    King6997 9:35:12 pm

    but a>=7

    binmu 9:35:12 pm

    but a must be greater than 7

    zheng 9:35:12 pm

    hoever a>=7

    nsato 9:35:14 pm

    But a is at least 7, which is a contradiction.

    nsato 9:35:21 pm

    Therefore, M contains an infinite number of primes.

    nsato 9:35:27 pm

    We're actually very close to the end now.

    nsato 9:35:32 pm

    nsato 9:36:06 pm

    We want to show that p is a factor of some product, minus 1. Does this remind us of any results?

    nsato 9:37:28 pmWe could try using Fermat's Little Theorem.

    nsato 9:37:37 pm

    nsato 9:38:37 pm

    nsato 9:39:04 pm

    Binomial-theorem 9:39:36 pm

    if p_1=p_2=...p_k=a(mod p)?

    AndroidFusion 9:39:36 pm

    let k = p-1 and p1, p2, ... pk be congruent mod p

    nsato 9:39:48 pm

    nsato 9:40:12 pm

    The primes p_1, p_2, ..., p_{p - 1} must also be distinct from p. But if p was in M, then this argument would not be necessary in the

    first place.

    nsato 9:40:50 pm

    How do we know that there exist p - 1 primes in M that are congruent modulo p?

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    Dunedubby 9:41:38 pm

    then use pigeonhole principle - with infinite pigeons

    AndroidFusion 9:41:38 pm

    there are an infinite number of primes, so use pidgeon principle

    Dunedubby 9:41:38 pm

    by the pigeonhole principle, with infinite pigeons, then at least p-1 must be in some a mod p (where a /== 0)

    aleph0 9:41:38 pm

    there are an infinite number, so by pigeonhole

    AndroidFusion 9:41:38 pm

    *pigeonhole principle

    mjseaman1 9:41:52 pm

    M has infinitely many primes, so the pigeonhole principle works

    nsato 9:41:53 pm

    We have shown that the set M is infinite.

    nsato 9:42:00 pm

    Hence, if we reduce every element in M modulo p, then some nonzero residue must appear an infinite number of times.

    nsato 9:42:16 pm

    Then we simply take p - 1 primes p_1, p_2, ..., p_{p - 1} that have this residue.

    nsato 9:42:35 pm

    Therefore, M contains every prime p.

    nsato 9:42:49 pm

    SUMMARY

    nsato 9:42:51 pm

    In today's class, we saw how to effectively use Fermat's Little Theorem and Euclid's Theorem. We also saw the power of algebra in

    number theory problems. We can solve many number theory problems effectively by putting a number in the right form (like

    expressing an even number in the form 2n), and using factorization. When you see a number theory problem, convert the words

    into mathematics, and let the equations do the work.

    nsato 9:43:06 pm

    Finally, we also saw how to solve problems using argument by contradiction. If a problem asks you to prove that something cannot

    occur, this is usually a good sign to use contradiction.

    nsato 9:43:14 pm

    That's it for today's class.

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    Art of Problem Solving

    WOOT

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    Class Transcript 01/21 - Modular Arithmetic B

    Valentin Vornicu 7:31:07 pm

    WOOT 2012-13: Modular Arithmetic B

    Valentin Vornicu 7:31:16 pm

    In today's lecture, we will continue our look at number theory and modular arithmetic. The first half of our class will focus on the use

    of Fermat's Little Theorem and Euler's Theorem. In the second half, we'll cover some more general number theory problems that

    are typical for this level.

    Valentin Vornicu 7:31:24 pm

    We'll start with a nice warm-up exercise.

    Valentin Vornicu 7:31:29 pm

    Valentin Vornicu 7:31:42 pm

    How can we find the missing digit?

    Cogswell 7:32:03 pmlook mod 9

    dinoboy 7:32:03 pm

    Use mod 9?

    matholympiad25 7:32:03 pm

    mod 9!

    apple.singer 7:32:03 pm

    mod 9?

    Valentin Vornicu 7:32:24 pm

    We can try to find the sum of the 9 digits. If all 10 digits were present, then their sum would be 0 + 1 + 2 + .. . + 9 = 45. However, we

    can't find the sum of the 9 digits without computing 2 29. How is mod 9 going to help us here?

    filetmignon821 7:33:21 pmsum of the digits mod 9 is the same as the number itself mod 9

    apple.singer 7:33:21 pm

    the sum of the digits is congruent to the number itself mod 9

    Cogswell 7:33:21 pm

    number = sum of digits mod 9

    sjaelee 7:33:21 pm

    sum of digits is number mod 9

    Valentin Vornicu 7:33:28 pm

    We know that a number and the sum of its digits are congruent modulo 9, so we can reduce 2^29 modulo 9. What do we get?

    BOBBOBBOB 7:34:02 pm

    2^3=8=-1mod9

    krmathcounts 7:34:02 pm

    32 mod 9

    filetmignon821 7:34:02 pm

    its the same as 2^5 mod 9, which is 5.

    antimonyarsenide 7:34:02 pm

    5

    krmathcounts 7:34:02 pm

    5 mod 9

    sjaelee 7:34:02 pm

    5

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    Valentin Vornicu 7:34:10 pm

    Valentin Vornicu 7:34:12 pm

    So what is the missing digit?

    Doink 7:35:06 pm

    4

    Cogswell 7:35:06 pm

    4

    apple.singer 7:35:06 pm

    4

    loquidyE 7:35:06 pm

    4

    Seedleaf 7:35:06 pm

    4

    jerrytang 7:35:06 pm

    4

    nikgran 7:35:06 pm

    4

    duketip10 7:35:06 pm

    4

    fireonice 7:35:06 pm

    4

    Valentin Vornicu 7:35:07 pm

    The missing digit must be congruent to -5 modulo 9, so the missing digit is 4.

    Valentin Vornicu 7:35:11 pm

    Indeed, 2^29 = 536870912. All digits are present except 4.

    Valentin Vornicu 7:35:20 pm

    Valentin Vornicu 7:35:29 pm

    How do we start?

    Doink 7:36:02 pm

    mod 1000

    msinghal 7:36:02 pm

    mod 1000

    yjhan96 7:36:02 pm

    mod 1000?

    Cogswell 7:36:02 pm

    mod 1000

    giratina150 7:36:02 pm

    mod 1000

    Valentin Vornicu 7:36:05 pm

    Finding the last three digits of a number is an indication to work modulo 1000.

    Valentin Vornicu 7:36:09 pm

    Valentin Vornicu 7:36:12 pm

    How can we reduce this?

    Cogswell 7:36:41 pm

    euler's thm

    Konigsberg 7:36:41 pm

    the euler's theorem

    filetmignon821 7:36:41 pm

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    phi(1000)=400, so we find 8^7 mod 400

    Konigsberg 7:36:41 pm

    it repeats every 400 times

    matholympiad25 7:36:41 pm

    we find 8^7 mod phi(1000)

    krmathcounts 7:36:41 pm

    phi(1000) = 400, and so 9^400 =1 mod 1000

    Valentin Vornicu 7:36:49 pm

    We can use Euler's Theorem.

    Valentin Vornicu 7:36:55 pm

    Valentin Vornicu 7:37:01 pm

    Valentin Vornicu 7:37:50 pm

    Probably the easiest way to compute this is to look at the first few powers of 8 modulo 400. We do this by starting with 1, then

    multiplying each term by 8 and reducing modulo 400.

    gurev 7:38:05 pm

    352

    Valentin Vornicu 7:38:09 pmThis gives us 1, 8, 64, 112, 96, 368, 144, 352.

    Valentin Vornicu 7:38:15 pm

    Valentin Vornicu 7:38:54 pm

    We have gone as far as we can with Euler's Theorem. We can try to compute this by looking at the first few powers of 9 modulo

    1000, but there is an easier way.

    Valentin Vornicu 7:38:58 pm

    Is there anything you notice about the number 9 that may make it easier to compute this power modulo 1000?

    Doink 7:39:40 pm

    9=10-1

    krmathcounts 7:39:40 pm

    -1 mod 10

    MathTwo 7:39:40 pm

    10-1

    fireonice 7:39:40 pm

    10-1

    yjhan96 7:39:40 pm

    10-1=9

    msinghal 7:39:40 pm

    9=10-1

    MathForFun 7:39:40 pm

    9==-1 (mod 10)

    Valentin Vornicu 7:40:09 pm

    Valentin Vornicu 7:40:17 pm

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    Valentin Vornicu 7:40:22 pm

    Therefore, the last three digits of 9^(8 7) are 081.

    Valentin Vornicu 7:40:55 pm

    Valentin Vornicu 7:41:18 pm

    We could verify that n^5 - n is divisible by 30 for n = 0, 1, 2, ..., 29. But what is an easier way?

    duketip10 7:42:38 pm

    do 2, 3, 5

    briantix 7:42:38 pm

    mod 2, 3, 5

    jerrytang 7:42:38 pm

    Take it mod 2, 3, and 5

    fireonice 7:42:38 pm

    factor and show it divides 2,3,5

    Cogswell 7:42:38 pm

    prove that it's divisible by 2, 3, and 5.

    msinghal 7:42:38 pmshow it is divisible for 2, 3, 5

    apple.singer 7:42:38 pm

    show it's divisible by 2, 3, and 5

    Valentin Vornicu 7:42:41 pm

    We use the prime factorization of 30 = 2 x 3 x 5. Hence, it suffices to show that n^5 - n is divisible by 2, 3, and 5, for all integers n.

    Valentin Vornicu 7:43:03 pm

    Is n 5 - n divisible by 2?

    ashgabat 7:43:12 pm

    Yes

    gurev 7:43:12 pm

    yes

    apple.singer 7:43:12 pm

    yes

    Konigsberg 7:43:12 pm

    yes it is

    MathForFun 7:43:12 pm

    yep

    pi.guy3.14 7:43:15 pm

    2 is c lear, if n is even, even-even=even, if its odd, odd-odd=even

    Valentin Vornicu 7:43:17 pm

    Yes. By reducing modulo 2, we can check for n = 0 and n = 1.

    Valentin Vornicu 7:43:22 pm

    Is n 5 - n divisible by 3?

    gurev 7:43:32 pm

    yes

    Konigsberg 7:43:32 pm

    yes

    fireonice 7:43:32 pm

    yes

    loquidyE 7:43:32 pm

    yes

    Doink 7:43:32 pm

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    yes

    pi.guy3.14 7:43:32 pm

    yes, plug in 0,1,2

    Valentin Vornicu 7:43:32 pm

    Yes. By reducing modulo 3, we can check for n = 0, 1, and 2.

    Valentin Vornicu 7:43:37 pm

    Is n 5 - n divisible by 5?

    pi.guy3.14 7:44:34 pm

    yes, plug in 0,1,2,3,4

    mrkidney 7:44:34 pm

    Yes, by Fermat's little theorem

    sjaelee 7:44:34 pm

    fermat's

    tareyza 7:44:34 pm

    yes

    duketip10 7:44:34 pm

    yes. n^4=1 by fermat's little theorem

    BOBBOBBOB 7:44:34 pm

    yep

    Valentin Vornicu 7:44:40 pmYes. By reducing modulo 5, we can check for n = 0, 1, 2, 3, and 4. We can also simply cite Fermat's Little Theorem for the prime p =

    5.

    Valentin Vornicu 7:44:46 pm

    Hence, n^30 - n is divisible by 2 x 3 x 5 = 30 for all integers n.

    Valentin Vornicu 7:44:51 pm

    This simple problem illustrates an important principle. If you want to prove that a number is divisible by n, it helps to look at the

    individual prime powers of n.

    Valentin Vornicu 7:45:06 pm

    Valentin Vornicu 7:45:20 pm

    We start by trying to reduce the given expression modulo 2000, but all the bases are already less than 2000. So what else can we

    do?

    fractals 7:46:11 pm

    take mod 16 and mod 125

    gurev 7:46:11 pm

    do mod 16 then mod 125

    Cogswell 7:46:11 pm

    split 2000 into 16 and 125

    apple.singer 7:46:11 pm

    factor 2000=2^4*5^3?

    aerrowfinn72 7:46:11 pm

    taking mod 125 and 16

    filetmignon821 7:46:11 pm

    2000=2^4*5^3, so we have to prove that our expression is divisible by 16 and 125.

    Valentin Vornicu 7:46:13 pm

    The prime factorization of 2000 is 2 4 x 5^3, so we look at the factors 2 4 = 16 and 5^3 = 125 separately.

    Valentin Vornicu 7:46:16 pm

    How does the given expression reduce modulo 16?

    jeff10 7:47:56 pm

    9^n-9^n+12^n-12^n

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    loquidyE 7:47:56 pm

    9^n - 9^n + 12^n - 12^n == 0 (mod 16)

    Seedleaf 7:47:56 pm

    12^n - (-4)^n

    filetmignon821 7:47:56 pm

    it comes out to (-7)^n-(-7)^n+(-4)^n-(-4)^n and everything cancels nicely.

    Valentin Vornicu 7:48:03 pm

    Valentin Vornicu 7:48:17 pm

    Therefore, 121^n - 25^n + 1900 n - (-4) n is divisible by 16.

    Valentin Vornicu 7:48:33 pm

    Modulo 125 we have a similar situation:

    Valentin Vornicu 7:48:40 pm

    Valentin Vornicu 7:48:45 pmTherefore, 121^n - 25^n + 1900 n - (-4) n is divisible by 125.

    Valentin Vornicu 7:49:23 pm

    We conclude that 121 n - 25^n + 1900 n - (-4)^n is divisible by 2000 for all positive integers n.

    Valentin Vornicu 7:49:31 pm

    Valentin Vornicu 7:50:24 pm

    What can we do with this equation?

    Cogswell 7:50:49 pm

    let's use the problem above: n^5-n (mod 30).

    loquidyE 7:50:49 pm

    check it for mod 2, 3, 5, etc.

    fireonice 7:50:49 pm

    take small mods

    Valentin Vornicu 7:50:54 pm

    We can use a previous problem and reduce it modulo 30. This gives us

    Valentin Vornicu 7:50:58 pm

    Valentin Vornicu 7:51:06 pm

    Now we need to find bounds on n to find its exact value. What's one obvious lower bound on n?

    Cogswell 7:52:25 pm

    133

    fireonice 7:52:25 pm

    133

    fractals 7:52:25 pm

    133

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    Seedleaf 7:52:25 pm

    133

    Konigsberg 7:52:25 pm

    133

    antimonyarsenide 7:52:25 pm

    n > 133

    jerrytang 7:52:25 pm

    133

    duketip10 7:52:25 pm

    133

    Valentin Vornicu 7:52:29 pm

    Clearly n > 133.

    Valentin Vornicu 7:52:33 pm

    Now we need to find an upper bound on n.

    Valentin Vornicu 7:53:22 pm

    The largest term in the left-hand side is 133^5, so we compare all the other terms to 133^5.

    Valentin Vornicu 7:53:25 pm

    We know that 110^5 < 133^5.

    Valentin Vornicu 7:53:28 pm

    Can we compare 84^5 + 27 5 and 133^5?

    Konigsberg 7:53:59 pm

    yes, it is smaller than 133^5

    pi.guy3.14 7:53:59 pm

    adding them up, they are still smaller

    aerrowfinn72 7:53:59 pm

    LHS is definitely smaller

    fractals 7:53:59 pm

    84 + 27 < 133, so the sum is less than 133^5

    filetmignon821 7:53:59 pm

    yes, 84^5+27^5

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    nikgran 7:56:32 pm

    4,5

    yjhan96 7:56:32 pm

    4^5 and 5^5

    Valentin Vornicu 7:56:34 pm

    We see that 5^5/4^5 = 3125/1024 is approximately 3. Even better, this ratio is slightly larger than 3, which means we can use this

    ratio to get an upper bound on n:

    Valentin Vornicu 7:57:46 pm

    Valentin Vornicu 7:57:54 pm

    Therefore, n < 5/4 x 133 = 166 + 1/4.

    Valentin Vornicu 7:57:58 pm

    So what is n?

    nikgran 7:58:17 pm

    we see n = 144 is the answer

    Konigsberg 7:58:17 pm

    144

    fractals 7:58:17 pm

    144

    briantix 7:58:17 pm

    144

    filetmignon821 7:58:17 pm

    144

    mrkidney 7:58:17 pm

    144

    matholympiad25 7:58:17 pm

    n is 144

    Valentin Vornicu 7:58:39 pm

    The only n satisfying 133 < n < 166 + 1/4, and n congruent to 24 modulo 30 is n = 144.

    Valentin Vornicu 8:03:38 pm

    We had a little situation with the server. Everything seems to be back to normal now.

    Valentin Vornicu 8:03:54 pm

    Valentin Vornicu 8:04:20 pm

    To make things easier, let s(N) denote the sum of the digits of N for a positive integer N. Then A = s(4444^4444) and B = s(A). Let C

    = s(B). Then we want to find C.

    Valentin Vornicu 8:04:32 pm

    Is there anything we can say about s(N)?

    Konigsberg 8:04:51 pm

    we get the modulo 9, because taking digit sums preserves the remainder when the number is divided by 9

    jeff10 8:04:51 pm

    I am thinking mod 9

    MathForFun 8:04:51 pm

    mod 9

    Cogswell 8:04:51 pm

    it is congruent to N (mod 9)

    Valentin Vornicu 8:04:54 pm

    We know that s(N) is congruent to N modulo 9.

    Valentin Vornicu 8:05:02 pm

    We can we say that C is congruent to 4444^4444 modulo 9.

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    Valentin Vornicu 8:05:06 pm

    How can we compute this residue?

    alex31415 8:06:13 pm

    4444=7 mod 9

    aerrowfinn72 8:06:13 pm

    take 4444 mod 9 first and then take 4444 mod phi(9)

    loquidyE 8:06:13 pm

    congruent to 7^4444

    jeff10 8:06:13 pm

    7^7 (mod 9)=7 (mod 9)

    Valentin Vornicu 8:06:16 pm

    Valentin Vornicu 8:06:19 pm

    How can we simplify this?

    Seedleaf 8:06:31 pm

    eulers theorem

    MathTwo 8:06:31 pm

    euler?

    filetmignon821 8:06:31 pm

    4444^6=1 (mod 9)

    filetmignon821 8:06:31 pm

    by euler's theorem, this is just 4444^4 mod 9

    fractals 8:06:34 pm

    phi(9) = 6

    Valentin Vornicu 8:06:52 pm

    We can use Euler's Theorem.

    Valentin Vornicu 8:06:57 pm

    Valentin Vornicu 8:07:33 pm

    Valentin Vornicu 8:07:43 pm

    Valentin Vornicu 8:07:47 pm

    Therefore, C is congruent to 7 modulo 9.

    Valentin Vornicu 8:07:50 pm

    How can we find the exact value of C?

    Cogswell 8:08:48 pm

    bounding

    aerrowfinn72 8:08:48 pm

    bound it

    Valentin Vornicu 8:08:50 pm

    Let's proceed by finding an upper bound on A, which is the sum of the digits of 4444^4444. What is an upper bound on A?

    filetmignon821 8:09:44 pm

    we know that s(n) is less than or equal to 9(log n+1), so we have that s(4444^4444) is at most 9(4444log4444+1)

    Valentin Vornicu 8:09:49 pm

    We can try to find the exact number of digits in 4444^4444, but that would involve using logarithms. What if we didn't have a

    calculator? Can we find an estimate on A, even a crude one?

    nikgran 8:10:39 pm

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    Valentin Vornicu 8:14:02 pm

    We have shown that C is congruent to 7 modulo 9, so we conclude that C = 7.

    Valentin Vornicu 8:14:48 pm

    Valentin Vornicu 8:14:59 pm

    What makes this sum difficult to deal with?

    Cogswell 8:16:08 pm

    absolute values

    filetmignon821 8:16:08 pm

    absolute value

    alex31415 8:16:08 pm

    Absolute value signs

    matholympiad25 8:16:08 pm

    The absolute values!

    MathForFun 8:16:08 pm

    absolute values

    Konigsberg 8:16:08 pm

    because it is absolute values

    Valentin Vornicu 8:16:27 pm

    The absolute value signs make this sum difficult to deal with. How can we deal with the absolute value signs? Is there a convenient

    way of getting rid of them?

    lightbluemathangel 8:17:39 pm

    a_i > b_i for all i

    antimonyarsenide 8:17:39 pm

    WLOG let ai>bi

    gurev 8:17:39 pm

    We can switch the order of the pairs to make ai the bigger one, and the absolute values dissappear

    Porteradams 8:17:39 pm

    have ai always greater than or equal to bi

    Valentin Vornicu 8:17:42 pm

    Valentin Vornicu 8:18:01 pm

    Valentin Vornicu 8:18:11 pmEach summand in this is 1 or 6. How can we use this information?

    Valentin Vornicu 8:19:08 pm

    There are 999 summands, so let k of them be equal to 1, so the remaining 999 - k are equal to 6.

    Valentin Vornicu 8:19:19 pm

    Valentin Vornicu 8:19:23 pm

    Doink 8:20:46 pm

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    they are all in the range from 1 to 1998

    jeff10 8:20:46 pm

    They are all less than or equal to 1998

    lightbluemathangel 8:20:46 pm

    they're all from 1 to 1998

    Seedleaf 8:20:46 pm

    they are distinct and belong to the set 1,2,...1998

    Valentin Vornicu 8:20:48 pm

    Valentin Vornicu 8:21:24 pm

    So can we turn this fact into a related sum?

    jerrytang 8:22:12 pm

    The sum is 1997001

    giratina150 8:22:12 pm

    sum of (a_i)+sum of (b_i)=1998(1999)/2

    filetmignon821 8:22:12 pm

    they sum to whatever the sum of 1 through 1998 is

    krmathcounts 8:22:12 pm

    a1+b1+a2+b2+.....+a999+b999 = 1998*1999/2

    Valentin Vornicu 8:22:52 pm

    Valentin Vornicu 8:22:54 pm

    What can we do with these two equations?

    Cogswell 8:23:08 pm

    add 'em?

    lightbluemathangel 8:23:08 pmadd them and then divide by 2 to get the sum of the a_i's

    krmathcounts 8:23:08 pm

    add

    Valentin Vornicu 8:23:12 pm

    Valentin Vornicu 8:23:18 pm

    We know that the left-hand side is even. Therefore, 5k must be odd, which means k must be odd.

    Valentin Vornicu 8:23:21 pm

    Let k = 2n + 1 for some integer n.

    Valentin Vornicu 8:23:25 pm

    Now what?

    numbertheorist17 8:24:15 pm

    your done!

    Cogswell 8:24:15 pm

    sum=5994-5(2n+1)=5989-10n

    nikgran 8:24:15 pm

    we are done, the sum ends in 9

    pi.guy3.14 8:24:15 pm

    replace k with n

    Valentin Vornicu 8:24:17 pm

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    We can substitute into the equation above, to get

    Valentin Vornicu 8:24:21 pm

    Valentin Vornicu 8:24:26 pm

    It looks like we can conclude that this number always ends in 9, but what if n is sufficiently high so that 5989 - 10n is negative? For

    example, if n = 600, then 5989 - 10n = -11, which ends in 1.

    briantix 8:25:37 pm

    all the terms are positive

    ajb 8:25:37 pm

    Absolute values ensure it is posit ive

    apple.singer 8:25:37 pm

    absolute values. can't be negative

    gurev 8:25:37 pm

    How can this be negative there's absolute values?

    Valentin Vornicu 8:25:38 pm

    The original sum was expressed as a sum of absolute values, which must be nonnegative. And a nonnegative number of the form

    5989 - 10n must end in 9, so we are home free.

    Valentin Vornicu 8:25:47 pm

    Valentin Vornicu 8:25:53 pm

    How can we proceed?

    Cogswell 8:27:21 pm

    assume the contrary

    aerrowfinn72 8:27:21 pm

    go for proof by contradiction

    numbertheorist17 8:27:21 pm

    contradiction

    fractals 8:27:21 pm

    assume that you can then find contradiction

    Valentin Vornicu 8:27:22 pm

    We can argue by contradiction.

    Valentin Vornicu 8:27:44 pm

    Note that 2 x 5 - 1 = 9, 2 x 13 - 1 = 25, and 5 x 13 - 1 = 64 are all perfect squares.

    Valentin Vornicu 8:27:48 pm

    Suppose 2d - 1 = x^2, 5d - 1 = y^2, and 13d - 1 = z^2, for some positive integers x, y, and z.

    Valentin Vornicu 8:27:56 pm

    What does the first equation 2d - 1 = x^2 tell us?

    gurev 8:28:41 pmx is odd

    jerrytang 8:28:41 pm

    x is odd

    lightbluemathangel 8:28:41 pm

    x^2==1 (mod 2)

    giratina150 8:28:41 pm

    x is odd

    pi.guy3.14 8:28:41 pm

    x is odd

    Valentin Vornicu 8:28:44 pm

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    Subtract for difference of squares.

    ajb 8:31:18 pm

    subtract them and use difference of squares

    Valentin Vornicu 8:31:22 pm

    We can factor the right-hand side, to get 2d = (v + u)(v - u).

    Valentin Vornicu 8:31:27 pm

    Can we say anything interesting about the factors v + u and v - u?

    aerrowfinn72 8:31:43 pm

    same parity

    giratina150 8:31:43 pm

    same parity

    filetmignon821 8:31:43 pm

    they have the same parity

    fractals 8:31:43 pm

    must be of same parity

    Cogswell 8:31:43 pm

    same parity-->d is even. But d is odd.

    duketip10 8:31:43 pm

    same parity

    jerrytang 8:31:43 pmSame parity

    Valentin Vornicu 8:31:46 pm

    Either v + u and v - u are both even or both odd (since their difference 2u is an even number).

    Valentin Vornicu 8:31:51 pm

    If v + u and v - u are both even, then their product (v + u)(v - u) is a multiple of 4. If v + u and v - u are both odd, then their product

    (v + u)(v - u) is odd. In particular, (v + u)(v - u) cannot be equal to 2d, because d is odd.

    Valentin Vornicu 8:31:56 pm

    We have obtained a contradiction, so at least one of 2d - 1, 5d - 1, and 13d - 1 is not a perfect square.

    Valentin Vornicu 8:32:03 pm

    It can seem simple, but using parity can be a powerful tool. And whenever you find that a number is even (odd resp.), it is usually a

    good idea to write it in the form 2n (2n + 1 resp.).

    Valentin Vornicu 8:38:45 pm

    Valentin Vornicu 8:38:55 pm

    Let f(n) = 19 x 8^n + 17. What approach can we take here?

    Seedleaf 8:40:06 pm

    look at small values of n first

    fractals 8:40:06 pm

    find some mod that works

    filetmignon821 8:40:06 pm

    prove it's divisible by something

    ajb 8:40:06 pm

    Modulo small numbers

    Valentin Vornicu 8:40:10 pm

    We can see if there exists a number d > 1 such that f(n) = 19 x 8^n + 17 is always divisible by d, by reducing it modulo d.

    Valentin Vornicu 8:40:15 pm

    What numbers d should we try?

    pi.guy3.14 8:41:33 pm

    9

    BOBBOBBOB 8:41:33 pm

    9

    gurev 8:41:33 pm

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    9?

    Valentin Vornicu 8:41:35 pm

    We should try numbers d such that 8 n modulo d simplifies.

    Valentin Vornicu 8:41:40 pm

    The first obvious number to try is d = 8.

    Valentin Vornicu 8:41:52 pm

    Valentin Vornicu 8:42:10 pm

    (9 is a good idea because 8 = - 1 (mod 9))

    Valentin Vornicu 8:43:11 pm

    So reducing modulo 7 doesn't get us anywhere either.

    Valentin Vornicu 8:43:29 pm

    Valentin Vornicu 8:43:48 pm

    We can look at the first few values of f(n), and see which divisors come up.

    Valentin Vornicu 8:43:58 pm

    What is f(0) and its prime factorization?

    fireonice 8:45:03 pm

    36=2^2*3^2

    MathForFun 8:45:03 pm

    2^2 3^2

    Porteradams 8:45:03 pm

    2^2(3^2)

    loquidyE 8:45:03 pm

    36 = 2^2 * 3^2

    nikgran 8:45:03 pm

    2^2*3^2

    Seedleaf 8:45:03 pm

    36

    Valentin Vornicu 8:45:04 pm

    f(0) = 36 = 2^2 x 3^2.

    Valentin Vornicu 8:45:08 pm

    What is f(1) and its prime factorization?

    jerrytang 8:46:28 pm

    13^2; 169

    fireonice 8:46:28 pm

    169= 13^2

    msinghal 8:46:28 pm

    169=13^2

    duketip10 8:46:28 pm

    169=13^2

    yjhan96 8:46:28 pm

    169=13^2

    Valentin Vornicu 8:46:29 pm

    f(1) = 169 = 13^2.

    Valentin Vornicu 8:46:38 pm

    What is f(2) and its prime factorization?

    filetmignon821 8:47:15 pm

    1233=3^2*137

    Seedleaf 8:47:15 pm

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    1233, 9*137

    giratina150 8:47:15 pm

    3^2*137

    Valentin Vornicu 8:47:16 pm

    f(2) = 1233 = 3^2 x 137.

    Valentin Vornicu 8:47:20 pm

    What is f(3) and its prime factorization?

    filetmignon821 8:48:00 pm

    9745=5*1949

    giratina150 8:48:00 pm

    5*1949

    Valentin Vornicu 8:48:01 pm

    f(3) = 9745 = 5 x 1949.

    Valentin Vornicu 8:48:32 pm

    Do we see anything promising?

    msinghal 8:49:55 pm

    for even n, 9 is a factor. for n=1 mod 4, 13 is a factor

    ajb 8:49:55 pm

    Okay, so it doesn't seem there is an n which divides all f(x), but 3^2=9 divides every even one

    gurev 8:49:55 pmIf n is even, this is divisible by 3, which is apparent when the expression is taken mod 9.

    fractals 8:49:55 pm

    if n even divisible by 9

    dinoboy 8:49:55 pm

    every f(n) is divisible by a "small prime"

    Valentin Vornicu 8:49:58 pm

    We see that 3 divides both f(0) and f(2), so let's work modulo 3 and see what happens.

    Valentin Vornicu 8:50:06 pm

    Valentin Vornicu 8:50:07 pm

    When is 2^n + 2 divisible by 3?

    briantix 8:50:38 pm

    when n is even

    fractals 8:50:38 pm

    n even

    Doink 8:50:38 pm

    when n is even

    Seedleaf 8:50:38 pm

    when n is even

    Valentin Vornicu 8:50:45 pm

    Since 2 is congruent to -1 modulo 3, 2^n + 2 is divisible by 3 if and only if n is even, which means f(n) is divisible by 3 if and only if n

    is even.

    Valentin Vornicu 8:50:54 pm

    Now we must look at f(n) for odd n. Let n = 2t + 1. What modulus can we work with?

    gurev 8:51:31 pm

    5

    Konigsberg 8:51:31 pm

    5?

    Cogswell 8:51:33 pm

    5

    Valentin Vornicu 8:51:34 pm

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    Since 5 is a factor of f(3), let's work modulo 5.

    Valentin Vornicu 8:51:43 pm

    Valentin Vornicu 8:51:53 pm

    When is 4^t + 1 divisible by 5?

    briantix 8:52:25 pm

    when t is odd

    lightbluemathangel 8:52:25 pm

    t is odd

    yjhan96 8:52:25 pm

    when t is odd

    Konigsberg 8:52:25 pm

    when t is an odd number

    fireonice 8:52:25 pm

    when t is odd

    Valentin Vornicu 8:52:28 pm

    Since 4 is congruent to -1 modulo 5, 4^t + 1 is divisible by 5 if and only if t is odd.

    Valentin Vornicu 8:52:41 pm

    Now we must look at f(2t + 1) for t even. Let t = 2u. Then f(2t + 1) = f(4u + 1).

    Valentin Vornicu 8:52:44 pm

    What modulus can we work with?

    Seedleaf 8:52:59 pm

    13

    yjhan96 8:52:59 pm

    13

    gurev 8:52:59 pm

    13

    Cogswell 8:52:59 pm

    13?

    fireonice 8:52:59 pm

    13

    Valentin Vornicu 8:53:01 pm

    There's really only one choice. Since f(1) = 13 2, let's work modulo 13.

    Valentin Vornicu 8:53:09 pm

    Valentin Vornicu 8:53:28 pm

    Thus, f(4u + 1) is always divisible by 13.

    Valentin Vornicu 8:53:33 pm

    To summarize, if n is even, then f(n) is divisible by 3 (and f(n) > 3),

    Valentin Vornicu 8:53:38 pm

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    if n is of the form 4k + 3, then f(n) is divisible by 5 (and f(n) > 5),

    Valentin Vornicu 8:53:43 pm

    and if n is of the form 4k + 1, then f(n) is divisible by 13 (and f(n) > 13).

    Valentin Vornicu 8:53:49 pm

    Therefore, f(n) is composite for all nonnegative integers n.

    Valentin Vornicu 8:54:06 pm

    This problem shows that sometime you must try different moduli, even within the same problem, so don't be afraid to experiment.

    Valentin Vornicu 8:54:25 pm

    Valentin Vornicu 8:54:34 pm

    You may recall Euclid's proof that there are an infinite number of primes. How can we start?

    briantix 8:54:54 pm

    hmmm...contradiction?

    MathForFun 8:54:54 pm

    contradiction

    fireonice 8:54:54 pm

    assume there are a finite number of primes

    Porteradams 8:54:54 pm

    assume the contradiction

    jerrytang 8:54:54 pm

    Contradiction

    Valentin Vornicu 8:54:55 pm

    We argue by contradiction.

    Valentin Vornicu 8:55:01 pm

    Valentin Vornicu 8:55:07 pm

    Then what?

    Seedleaf 8:56:22 pm

    multiply them together

    Porteradams 8:56:22 pm

    multiply all p together

    giratina150 8:56:22 pm

    multiply together

    pi.guy3.14 8:56:22 pm

    product+!

    MathForFun 8:56:22 pm

    multiply them all!

    Valentin Vornicu 8:56:26 pm

    Valentin Vornicu 8:57:11 pm

    What can we say about P?

    fractals 8:57:37 pm

    it is 2 mod 3 if k is even

    Seedleaf 8:57:37 pm

    either P = 0 or 2 mod 3

    Valentin Vornicu 8:57:40 pm

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    Valentin Vornicu 8:58:28 pm

    Suppose P reduces to 2 modulo 3. Then what can we say about P?

    Konigsberg 8:59:28 pm

    then it is in the form of 3n+2

    Porteradams 8:59:28 pm

    it is of the form 3n+2 which was not on the list

    giratina150 8:59:28 pm

    it's of the form 3n+2

    jerrytang 8:59:28 pm

    It is of form 3n+2, new prime, contradiction

    briantix 8:59:28 pm

    it must have a prime factor that is 2 mod three

    Valentin Vornicu 8:59:30 pm

    All primes (other than 3) are congruent to 1 modulo 3 and 2 modulo 3. If P was only divisible by primes congruent to 1 modulo 3,

    then P itself would be congruent to 1 modulo 3. But P is congruent to 2 modulo 3, so P must be divisible by at least one prime that is

    congruent to 2 modulo 3.

    Valentin Vornicu 8:59:36 pm

    But by construction, P is relatively prime to all primes that are congruent to 2 modulo 3, contradiction.

    Valentin Vornicu 8:59:55 pm

    The only case that is left is if P reduces to 0 modulo 3. Unfortunately, there's no good way to handle this case. (We can divide out al

    the factors of 3, but there's nothing we can say about the resulting quotient.)

    Valentin Vornicu 8:59:58 pm

    Is there any way to fix this argument?

    filetmignon821 9:01:26 pm

    consider 3p_1p_2...p_k - 1. This must be divisible by some prime of the form 3n+2, but is not divisible by any of the p_i. this forces a

    contradiction.

    dinoboy 9:01:26 pm

    take P = 3p_1p_2...p_k + 2, then its forced to be 2 mod 3

    Valentin Vornicu 9:01:36 pm

    Valentin Vornicu 9:01:38 pmThen P always reduces to 2 modulo 3, and we can proceed as above.

    Valentin Vornicu 9:01:44 pm

    More generally, if a and b are relatively prime positive integers, then there are infinitely many primes of the form an + b. This result

    is known as Dirichlet Theorem, but it is an advanced result and is difficult to prove.

    Valentin Vornicu 9:01:49 pm

    The ideas in Euclid's proof lead to the next few problems.

    Valentin Vornicu 9:01:58 pm

    Valentin Vornicu 9:03:08 pm

    To get a feel for the problem, let's find the first few terms of the sequence.

    Valentin Vornicu 9:03:14 pm

    What is p_2?

    BOBBOBBOB 9:03:22 pm

    3

    lightbluemathangel 9:03:22 pm

    3

    matholympiad25 9:03:22 pm

    3

    filetmignon821 9:03:22 pm

    3

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    gurev 9:03:22 pm

    3

    aerrowfinn72 9:03:22 pm

    3

    Valentin Vornicu 9:03:23 pm

    p_2 is the largest prime divisor of p_1 + 1 = 3, so p_2 = 3.

    Valentin Vornicu 9:03:25 pm

    What is p_3?

    giratina150 9:03:33 pm

    7

    filetmignon821 9:03:33 pm

    7

    aerrowfinn72 9:03:33 pm

    7

    loquidyE 9:03:33 pm

    7

    fireonice 9:03:33 pm

    7

    Seedleaf 9:03:33 pm

    7

    Valentin Vornicu 9:03:33 pm

    p_3 is the largest prime divisor of p_1 p_2 + 1 = 7, so p_3 = 7.

    Valentin Vornicu 9:03:36 pm

    What is p_4?'

    BOBBOBBOB 9:03:42 pm

    43

    aerrowfinn72 9:03:42 pm

    43

    matholympiad25 9:03:42 pm

    43

    ajb 9:03:42 pm43

    giratina150 9:03:42 pm

    43

    fireonice 9:03:42 pm

    43

    Cogswell 9:03:42 pm

    43

    Valentin Vornicu 9:03:43 pm

    p_4 is the largest prime divisor of p_1 p_2 p_3 + 1 = 43, so p_4 = 43.

    Valentin Vornicu 9:03:45 pm

    What is p_5?

    giratina150 9:04:34 pm

    139?

    Cogswell 9:04:34 pm

    139

    Valentin Vornicu 9:04:35 pm

    p_5 is the largest prime divisor of p_1 p_2 p_3 p_4 + 1 = 1807 = 13 x 139, so p_4 = 139.

    Valentin Vornicu 9:04:42 pm

    We must show that 5 never appears in the sequence.

    Valentin Vornicu 9:04:44 pm

    How can we proceed?

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    saphireflame 9:05:58 pm

    proof by contradiction

    krmathcounts 9:05:58 pm

    assume 5 does show up?

    Valentin Vornicu 9:06:01 pm

    We can argue by contradiction.

    Valentin Vornicu 9:06:11 pm

    Valentin Vornicu 9:06:41 pm

    What does that say about this expression?

    Cogswell 9:06:57 pm

    p_1p_2/dotsp_n-1+1 = 2^Q3^R5^S

    MathTwo 9:06:57 pm

    some multiplication of 2s, 3 and 5

    briantix 9:06:57 pm

    so its 2^a*3^b*5^c

    jerrytang 9:06:57 pm

    Only primes are 2, 3, 5

    Valentin Vornicu 9:06:59 pm

    Valentin Vornicu 9:07:02 pm

    Can we say more?

    filetmignon821 9:08:25 pm

    this can't be a multiple of 2 or 3, since p_1 and p_2 are 2 and 3, so it must be a power of 5.

    Seedleaf 9:08:25 pm

    its only a power of 5

    fireonice 9:08:25 pm

    5 only because 2 and 3 show up in the sequence

    apple.singer 9:08:32 pm

    just 5, as 2 and 3 are p_1 and p_2 and thus will not divide the product +1

    Valentin Vornicu 9:08:35 pm

    Valentin Vornicu 9:08:45 pm

    Valentin Vornicu 9:09:01 pm

    Valentin Vornicu 9:09:10 pm

    Valentin Vornicu 9:09:12 pm

    What can we do with this equation?

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    fractals 9:10:18 pm

    mod 4

    briantix 9:10:18 pm

    factor it

    lightbluemathangel 9:10:18 pm

    factor out (5-1)

    giratina150 9:10:18 pm

    mod 4

    ajb 9:10:18 pm

    5^k-1=0 mod 4

    briantix 9:10:18 pm

    both sides must have a 2^2 in them but that's impossible for the left one

    filetmignon821 9:10:18 pm

    difference of kth powers forces p_1p_2...p_{n-1} to be divisible by 4

    aerrowfinn72 9:10:18 pm

    take it mod 4

    Valentin Vornicu 9:10:24 pm

    Valentin Vornicu 9:10:27 pm

    This equation tells us that 4 must be a factor of the left-hand side.

    Valentin Vornicu 9:10:34 pm

    But this is impossible because p_1 = 2 and all further terms p_2, ..., p_{n - 1} are odd.

    Valentin Vornicu 9:10:37 pm

    We have a contradiction, so 5 cannot be a member of this sequence.

    Valentin Vornicu 9:10:49 pm

    I'll let you think about why 11 is also not in the sequence.

    Valentin Vornicu 9:10:55 pm

    Valentin Vornicu 9:11:27 pm

    I actually proposed this problem for the IMO TST. Only the 6 students that actually made the team managed to solve it.

    Valentin Vornicu 9:11:48 pm

    Let's start by looking at small primes. Is the prime 2 in P?

    filetmignon821 9:12:30 pm

    i bet it is, since P is the set of all primes

    msinghal 9:12:30 pm

    yes, all primes are in P

    Cogswell 9:12:30 pm

    yes, since it 's prime

    Valentin Vornicu 9:12:36 pm

    That's not exactly a proof

    gurev 9:13:13 pm

    Did you mean is it in M, which is also true?

    Valentin Vornicu 9:13:30 pm

    Oops, a small typo. We're looking at 2 being in M.

    Cogswell 9:14:06 pm

    let A be the set of all odd primes in M. The result follows.

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    dinoboy 9:14:06 pm

    suppose not, take an odd prime in M and let it be A. Then we reach a contradiction so 2 is in M

    Valentin Vornicu 9:14:08 pm

    Since M contains at least three primes, there is an odd prime in M, say p. Let A = {p}. Then by the condition in the problem, all the

    prime factors of p - 1 are also in M.

    Valentin Vornicu 9:14:39 pm

    But p - 1 is even, so 2 is in M.

    Valentin Vornicu 9:14:48 pm

    Is the prime 3 in M?

    gurev 9:15:40 pm

    Yes

    Cogswell 9:15:40 pm

    yes

    Valentin Vornicu 9:15:47 pm

    I need a justification to give you all the 7 points.

    filetmignon821 9:17:09 pm

    if we have another prime p in M that is 1 mod 3, then 3 is in M. If p is 2 mod 3, then 2p is 1 mod 3, so 2p-1 is divisible by three. thus, 3 is in M

    ajb 9:17:09 pm

    there must be at least two elements equal to 1 mod 3 or 2 mod 3 by pigeonhole

    Cogswell 9:17:09 pm

    take the two primes in M that are not 3. If one is 1 mod 3, the result follows. If they both are 2 mod 3, their product is 1 mod 3, and the resultfollows

    Valentin Vornicu 9:17:22 pm

    If M contains a prime of the form p = 3n + 1, then take A = {p}. Then all the factors of p - 1 = 3n are also in M, so 3 is in M.

    Valentin Vornicu 9:17:28 pm

    Otherwise, M contains two primes of the form 3n + 2, say p and q, and we can take A = {p, q}. Then all the prime factors of pq - 1

    are also in M.

    Valentin Vornicu 9:17:31 pm

    But pq - 1 is congruent to 2 x 2 - 1 = 3, or 0 modulo 3, so pq - 1 is divisible by 3. Therefore, 3 is in M. In either case, 3 is in M.

    Valentin Vornicu 9:17:35 pm

    Is the prime 5 in M?

    filetmignon821 9:19:05 pmyeah. 2*3-1=5

    ajb 9:19:05 pm

    yes, since 2*3=6, and 6-1=5

    loquidyE 9:19:05 pm

    yes; 2*3 - 1 = 5

    msinghal 9:19:05 pm

    yes, 2*3 - 1

    Valentin Vornicu 9:19:06 pm

    We know 2 and 3 are in M, so take A = {2, 3}. Then 2 x 3 - 1 = 5, so 5 is in M.

    Valentin Vornicu 9:19:11 pm

    Is the prime 7 in M?

    briantix 9:20:00 pm

    3*5-1=2*7 so yes

    apple.singer 9:20:00 pm

    yes, take {3,5}

    antimonyarsenide 9:20:00 pm

    yes, 7 | 3*5 - 1

    gurev 9:20:00 pm

    Yeah, 3*5-1

    Seedleaf 9:20:00 pm

    3*5 - 1 = 2*7

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    Cogswell 9:20:00 pm

    yes (a={3,5})

    filetmignon821 9:20:00 pm

    yeah, 3*5-1=14, which is divisible by 7

    Valentin Vornicu 9:20:01 pm

    We know 3 and 5 are in M, so take A = {3, 5}. Then 3 x 5 - 1 = 14 = 2 x 7, so 7 is in M.

    Valentin Vornicu 9:20:20 pm

    Hence, the first few primes are in M. We want to prove that M contains all primes.

    Valentin Vornicu 9:20:25 pm

    What intermediate result can we try proving first?

    gurev 9:20:43 pm

    infinitely many

    antimonyarsenide 9:20:43 pm

    M contains infinitely many primes

    Valentin Vornicu 9:20:45 pm

    Let's try proving first that M is infinite.

    Valentin Vornicu 9:20:50 pm

    Valentin Vornicu 9:21:06 pm

    As a start, let's take A to be all the primes in M, except one, say p_i. (Remember that A must be a proper subset of M.)

    Valentin Vornicu 9:21:11 pm

    Valentin Vornicu 9:21:19 pm

    Cogswell 9:22:29 pm

    the only prime that can divide it is p_i

    NewAlbionAcademy 9:22:29 pm

    It isn't divisible by any of them, so it must be a power of p_i

    gurev 9:22:29 pm

    it's a power of pi

    Valentin Vornicu 9:22:47 pm

    Valentin Vornicu 9:23:02 pm

    Valentin Vornicu 9:23:17 pm

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    Valentin Vornicu 9:23:31 pm

    To make these equations simpler to work with, we start by omitting the smallest primes.

    Valentin Vornicu 9:23:38 pm

    Valentin Vornicu 9:25:06 pm

    Valentin Vornicu 9:25:37 pm

    Valentin Vornicu 9:25:41 pm

    Valentin Vornicu 9:26:15 pm

    What can we do with these two expressions for P?

    giratina150 9:26:33 pm

    equate

    Cogswell 9:26:33 pm

    set them equal to each other

    Porteradams 9:26:33 pm

    set them equal

    fireonice 9:26:33 pm

    set them equal to each other

    Valentin Vornicu 9:26:35 pm

    We can equate them.

    Valentin Vornicu 9:26:45 pm

    Valentin Vornicu 9:26:54 pm

    Let's start analyzing this equation.

    Valentin Vornicu 9:27:02 pm

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    Valentin Vornicu 9:27:04 pm

    Also, P is at least 2 x 3 x 5 x 7. So what can we say about a?

    Cogswell 9:28:06 pm

    a is at least 6

    Porteradams 9:28:06 pm

    at least 6

    pi.guy3.14 9:28:06 pm

    at least 6

    Valentin Vornicu 9:28:08 pm

    Valentin Vornicu 9:28:19 pm

    Valentin Vornicu 9:28:23 pm

    So what can we say about b?

    Cogswell 9:28:50 pm

    it is at least 1.

    Valentin Vornicu 9:28:52 pm

    Valentin Vornicu 9:28:59 pm

    Valentin Vornicu 9:29:14 pm

    matholympiad25 9:29:43 pm

    it is odd.

    fractals 9:29:43 pm

    odd

    MathForFun 9:29:43 pm

    It's odd

    giratina150 9:29:43 pm

    odd

    jerrytang 9:29:43 pm

    It is odd.

    fireonice 9:29:43 pm

    it is odd

    Valentin Vornicu 9:29:46 pm

    Valentin Vornicu 9:29:51 pm

    gurev 9:29:55 pm

    b is 1

    MathForFun 9:29:55 pm

    so b=1

    Valentin Vornicu 9:29:56 pm

    Therefore, b = 1.

    Valentin Vornicu 9:30:01 pm

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    Art of Problem Solving

    WOOT

    Copyright AoPS Incorporated. This page is copyrighted material. You can view and print this page for your own use, but you

    cannot share the contents of this file with others.

    Class Transcript 01/24 - Modular Arithmetic B

    Valentin Vornicu 9:00:37 pm

    WOOT 2012-13: Modular Arithmetic B

    Valentin Vornicu 9:00:44 pm

    In today's lecture, we will continue our look at number theory and modular arithmetic. The first half of our class will focus on the use

    of Fermat's Little Theorem and Euler's Theorem. In the second half, we'll cover some more general number theory problems that

    are typical for this level.

    Valentin Vornicu 9:00:49 pm

    We'll start with a nice warm-up exercise.

    Valentin Vornicu 9:00:55 pm

    Valentin Vornicu 9:01:22 pm

    How can we find the missing digit?

    KingSmasher3 9:01:37 pmmod 9!

    vishankjs 9:01:37 pm

    look mod 9

    v_Enhance 9:01:37 pm

    Take mod 9

    Valentin Vornicu 9:01:45 pm

    We can try to find the sum of the 9 digits. If all 10 digits were present, then their sum would be 0 + 1 + 2 + .. . + 9 = 45. However, we

    can't find the sum of the 9 digits without computing 2 29.

    Valentin Vornicu 9:01:52 pm

    We know that a number and the sum of its digits are congruent modulo 9, so we can reduce 2^29 modulo 9. What do we get?

    Mousie 9:04:57 pm5mod9

    greensteg2 9:04:57 pm

    5 mod 9

    Bryan.C 9:04:57 pm

    5

    aopsqwerty 9:04:57 pm

    5

    woodstock 9:04:57 pm

    5

    AkshajK 9:04:57 pm

    zhaoamc 9:04:57 pm

    5 mod 9

    AkshajK 9:04:57 pm

    mentalgenius 9:04:57 pm

    5

    Valentin Vornicu 9:05:01 pm

    Valentin Vornicu 9:05:43 pm

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    So what is the missing digit?

    baijiangchen 9:06:02 pm

    4

    ABCDE 9:06:02 pm

    so the digit is 4

    vishankjs 9:06:02 pm

    4

    Mousie 9:06:02 pm

    4?

    zhaoamc 9:06:02 pm

    4

    woodstock 9:06:02 pm

    4

    sdaops 9:06:02 pm

    4

    cire_il 9:06:02 pm

    4

    Valentin Vornicu 9:06:03 pm

    The missing digit must be congruent to -5 modulo 9, so the missing digit is 4.

    Valentin Vornicu 9:06:06 pmIndeed, 2^29 = 536870912. All digits are present except 4.

    Valentin Vornicu 9:06:13 pm

    Valentin Vornicu 9:06:19 pm

    How do we start?

    algebra1337 9:07:42 pm

    mod 1000

    KingSmasher3 9:07:42 pm

    last three digits means mod 1000

    hawqish 9:07:42 pm

    take mod 1000

    maxmk 9:07:42 pm

    mod 1000

    aopsqwerty 9:07:42 pm

    mod 1000

    Valentin Vornicu 9:07:42 pm

    Finding the last three digits of a number is an indication to work modulo 1000.

    Valentin Vornicu 9:07:48 pm

    Valentin Vornicu 9:07:51 pm

    How can we reduce this?

    zhaoamc 9:08:00 pm

    eulor's totient thm

    sdaops 9:08:00 pm

    We want mod 1000, so phi(1000) is useful

    Porteradams 9:08:00 pm

    eulers theorem

    Valentin Vornicu 9:08:02 pm

    We can use Euler's Theorem.

    Valentin Vornicu 9:08:05 pm

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    Valentin Vornicu 9:08:47 pm

    Valentin Vornicu 9:08:52 pm

    Probably the easiest way to compute this is to look at the first few powers of 8 modulo 400. We do this by starting with 1, then

    multiplying each term by 8 and reducing modulo 400.

    Valentin Vornicu 9:08:56 pm

    This gives us 1, 8, 64, 112, 96, 368, 144, 352.

    Valentin Vornicu 9:09:04 pm

    Valentin Vornicu 9:09:11 pm

    We have gone as far as we can with Euler's Theorem. We can try to compute this by looking at the first few powers of 9 modulo

    1000, but there is an easier way.

    Valentin Vornicu 9:09:15 pm

    Is there anything you notice about the number 9 that may make it easier to compute this power modulo 1000?

    sdaops 9:09:44 pm

    It's 10-1

    greensteg2 9:09:44 pm

    9 = 10 - 1

    ksun48 9:09:44 pm

    oh... It's 10-1

    Mousie 9:09:44 pm

    10 - 1

    baijiangchen 9:09:44 pm

    9 = (10-1)

    GeorgiaTechMan 9:09:44 pm

    and its congruent to -1 mod 10!

    KingSmasher3 9:09:44 pm

    teethpaste 9:09:44 pm

    9=10-1?

    Valentin Vornicu 9:09:50 pm

    Valentin Vornicu 9:10:58 pm

    Valentin Vornicu 9:11:03 pm

    Therefore, the last three digits of 9^(8 7) are 081.

    Valentin Vornicu 9:11:11 pm

    Valentin Vornicu 9:11:19 pm

    We could verify that n^5 - n is divisible by 30 for n = 0, 1, 2, ..., 29. But what is an easier way?

    maxmk 9:12:40 pm

    2, 3, 5

    AkshajK 9:12:40 pm

    SkinnySanta 9:12:40 pm

    fermats little theorem for 2,3,5 and apply c rt

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    Yes. By reducing modulo 5, we can check for n = 0, 1, 2, 3, and 4. We can also simply cite Fermat's Little Theorem for the prime p =

    5.

    Valentin Vornicu 9:14:42 pm

    Hence, n^5 - n is divisible by 2 x 3 x 5 = 30 for all integers n.

    Valentin Vornicu 9:14:46 pm

    This simple problem illustrates an important principle. If you want to prove that a number is divisible by n, it helps to look at the

    individual prime powers of n.

    Valentin Vornicu 9:15:46 pm

    Valentin Vornicu 9:15:54 pm

    We start by trying to reduce the given expression modulo 2000, but all the bases are already less than 2000. So what else can we

    do?

    algebra1337 9:17:22 pm

    mod 16 and mod 125

    anwang16 9:17:22 pm

    mod 16? mod 25?

    aopsqwerty 9:17:22 pm

    show it is divisible by 16 and 125

    baijiangchen 9:17:22 pm

    factor into 2^4 5^3

    KingSmasher3 9:17:22 pm

    prove divisibility by 125 and 16

    GeorgiaTechMan 9:17:22 pm

    try mod 16 and mod 125

    Valentin Vornicu 9:17:24 pm

    The prime factorization of 2000 is 2 4 x 5^3, so we look at the factors 2 4 = 16 and 5^3 = 125 separately.

    Valentin Vornicu 9:17:28 pm

    How does the given expression reduce modulo 16?

    GeorgiaTechMan 9:18:53 pm

    9^n-9^n+(-4)^n - (-4)^n

    greensteg2 9:18:53 pm

    9^n - 9^n + 12^n - (-4)^n

    teethpaste 9:18:53 pm

    12^n - (-4)^n

    DVA6102 9:18:53 pm

    9^n-9^n+12^n-12^n=0. yes

    Mousie 9:18:53 pm

    12^n - (-4)^n

    Valentin Vornicu 9:18:54 pm

    Valentin Vornicu 9:18:57 pm

    Therefore, 121^n - 25^n + 1900 n - (-4) n is divisible by 16.

    Valentin Vornicu 9:19:06 pm

    How does the given expression reduce modulo 125?

    anwang16 9:20:12 pm

    (121)^n-(25)^n+25^n-121^n

    sdaops 9:20:12 pm

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    (-4)^n -25^n + 25^n - (-4)^n = 0

    Mousie 9:20:12 pm

    0 as well...terms cancel

    woodstock 9:20:12 pm

    121^n - 25^n + 25^n - 121^n

    Valentin Vornicu 9:20:15 pm

    Valentin Vornicu 9:20:17 pm

    Therefore, 121^n - 25^n + 1900 n - (-4) n is divisible by 125.

    Valentin Vornicu 9:20:21 pm

    We conclude that 121 n - 25^n + 1900 n - (-4)^n is divisible by 2000 for all positive integers n.

    Valentin Vornicu 9:20:31 pm

    Valentin Vornicu 9:20:42 pm

    What can we do with this equation?

    anwang16 9:22:11 pm

    mod something?

    Valentin Vornicu 9:22:18 pm

    What mod should we use?

    GeorgiaTechMan 9:22:31 pm

    actually, just 2, 3, and 5 suffice

    SkinnySanta 9:22:31 pm

    look modulo 30 since n^5 = n mof 30

    Valentin Vornicu 9:22:37 pm

    We can use a previous problem and reduce it modulo 30. This gives us

    Valentin Vornicu 9:22:40 pm

    Valentin Vornicu 9:22:45 pm

    Now we need to find bounds on n to find its exact value. What's one obvious lower bound on n?

    hawqish 9:24:28 pm

    > 133?

    ABCDE 9:24:28 pm

    133

    Mousie 9:24:28 pm

    oops 133

    greensteg2 9:24:28 pm

    133

    mentalgenius 9:24:28 pm

    133

    SkinnySanta 9:24:28 pm

    133

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    woodstock 9:24:28 pm

    133

    GeorgiaTechMan 9:24:28 pm

    n>133

    Valentin Vornicu 9:24:29 pm

    Clearly n > 133.

    Valentin Vornicu 9:24:32 pm

    Now we need to find an upper bound on n.

    Valentin Vornicu 9:24:37 pm

    The largest term in the left-hand side is 133^5, so we compare all the other terms to 133^5.

    Valentin Vornicu 9:25:21 pm

    We know that 110^5 < 133^5.

    Valentin Vornicu 9:25:26 pm

    Can we compare 84^5 + 27 5 and 133^5?

    woodstock 9:26:41 pm

    84^5 + 27^5 < 111^5 < 133^5

    Mousie 9:26:41 pm

    since 84^5 + 27^5 < 133^5

    AkshajK 9:26:41 pm

    (a+b)^5 > a^5 + b^5

    DVA6102 9:26:41 pm

    84^5+27^5 < 133^5

    Valentin Vornicu 9:26:42 pm

    Note that 84^5 + 27^5 < (84 + 27)^5 = 111^5 < 133^5.

    Valentin Vornicu 9:26:48 pm

    Valentin Vornicu 9:26:52 pm

    What we would like to do now is take the fifth root of both sides. The problem is we would get the fifth root of 3, and it's not clear

    what to do with this number.

    Valentin Vornicu 9:26:56 pmSo let's see if we can replace 3 by the ratio of two fifth powers.

    Valentin Vornicu 9:27:02 pm

    The first few fifth powers are 1 5 = 1, 2^5 = 32, 3^5 = 243, 4 5 = 1024, 5^5 = 3125, and 6^5 = 7776, and so on. Do we see any fifth

    powers that approximately have a ratio of 3?

    ABCDE 9:28:42 pm

    5^5 and 4^5

    aopsqwerty 9:28:42 pm

    4^5 and 5^5

    jhfrost314 9:28:42 pm

    5^5/4^5

    greensteg2 9:28:42 pm

    4^5 and 5^5

    pi314159265358979 9:28:42 pm

    5^5 and 4^5

    hawqish 9:28:42 pm

    5^4 and 4^5

    baijiangchen 9:28:42 pm

    (5/4)

    GeorgiaTechMan 9:28:42 pm

    4 and 5

    sdaops 9:28:42 pm

    Thus n < 133*5/4

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    zhaoamc 9:28:42 pm

    5^5/4^5

    Valentin Vornicu 9:28:43 pm

    We see that 5^5/4 5 = 3125/1024 is approximately 3.

    Valentin Vornicu 9:28:49 pm

    Even better, this ratio is slightly larger than 3, which means we can use this ratio to get an upper bound on n:

    Valentin Vornicu 9:28:52 pm

    Valentin Vornicu 9:28:55 pm

    Therefore, n < 5/4 x 133 = 166 + 1/4.

    Valentin Vornicu 9:28:58 pm

    So what is n?

    Minamoto 9:29:08 pm

    144

    AkshajK 9:29:08 pm

    144

    KingSmasher3 9:29:08 pm

    144

    bobcat1209:29:08 pm

    144

    DVA6102 9:29:08 pm

    144

    Valentin Vornicu 9:29:10 pm

    The only n satisfying 133 < n < 166 + 1/4, and n congruent to 24 modulo 30 is n = 144.

    Valentin Vornicu 9:30:25 pm

    Valentin Vornicu 9:30:33 pm

    To make things easier, let s(N) denote the sum of the digits of N for a positive integer N. Then A = s(4444^4444) and B = s(A). Let C

    = s(B). Then we want to find C.

    Valentin Vornicu 9:31:43 pm

    Is there anything we can say about s(N)?

    DVA6102 9:33:15 pm

    Can we take it mod 9, since s(x)==x mod 9?

    sdaops 9:33:15 pm

    Congruent to N mod 9

    maxmk 9:33:15 pm

    s(N) == N (mod 9)

    ksun48 9:33:15 pm

    9|N-s(N)

    ABCDE 9:33:15 pm

    s(N)==N mod 9

    SkinnySanta 9:33:15 pm

    s(n) = n mod 9

    Valentin Vornicu 9:33:15 pm

    We know that s(N) is congruent to N modulo 9.

    Valentin Vornicu 9:33:26 pm

    So we can we say that C is congruent to 4444^4444 modulo 9.

    Valentin Vornicu 9:33:30 pm

    How can we compute this residue?

    woodstock 9:35:06 pm

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    4444 = 7 (mod 9)

    Bryan.C 9:35:06 pm

    euler's theorem

    hawqish 9:35:06 pm

    use euler's theorem

    anwang16 9:35:06 pm

    Get rid of most of the numbers and end up with 7^4444 which is -2^4444

    AkshajK 9:35:06 pm

    Valentin Vornicu 9:35:10 pm

    Valentin Vornicu 9:35:18 pm

    We can use Euler's Theorem to simplify this.

    Valentin Vornicu 9:35:22 pm

    Valentin Vornicu 9:36:11 pm

    Valentin Vornicu 9:36:16 pm

    Valentin Vornicu 9:36:21 pm

    Therefore, C is congruent to 7 modulo 9.

    aopsqwerty 9:36:25 pm

    but 7=-2 (mod 9)

    Valentin Vornicu 9:36:40 pm

    True we could have used that, and it would have probably been a little faster. Same idea though!

    Valentin Vornicu 9:36:47 pm

    Now, how can we find the exact value of C?

    Minamoto 9:38:16 pm

    use bounds

    aopsqwerty 9:38:16 pm

    find bounds?

    SkinnySanta 9:38:16 pm

    find upper bounds based on the number of digits

    sdaops 9:38:16 pm

    Let's figure out roughly how big C is; if it's a single digit, we know it already.

    V