amc 8 at mmc - metroplex math circle · 8 years older than she really is, we would be the same age....

AMC 8 at MMC

Adrian and Titu Andreescu

November 2, 2013

Problems

1. Find all 3-digit numbers abc such that ab + bc + ca = abc. Note: In problems like this, a, b, care digits, i.e., 0 ≤ a, b, c ≤ 9, and we do not allow leading zeros.

2. Evaluate:

a = (128− 2)(128− 22)(128− 2

3) . . . (128− 2

7)(128− 2

8).

3. Show that (3201

+ 3204

)÷ (3201 − 3

200+ 3

199) is a perfect square.

4. I am 30 years older than my daughter. If I were 2 times younger than I really am and she was

8 years older than she really is, we would be the same age. How old are we?

5. Find all numbers of the form xyxz that are divisible by 11, 12, and 13.

6. Prove that

145678 + 456781 + 567814 + 678145 + 781456 + 814567

is the product of six different prime numbers.

7. Find the last digit of 21+2+3+...+2009

.

8. Find the number of zeros in which 50! ends.

9. The average cost of a long-distance call in 1985 was 41 cents per minute, and the average cost

of a long-distance call in 2005 was 7 cents per minute. Find the approximate percent decrease

in the cost per minute of a long-distance call.

10. An athlete’s target heart rate, in beats per minute, is 80% of the theoretical maximum heart

rate. The maximum heart rate is found by subtracting the athlete’s age, in years, from 220. To

the nearest whole number, what is the target heart rate of an athlete who is 26 years old? top

11. Miki has a dozen oranges of the same size and a dozen pears of the same size. Miki can extract

8 ounces of pear juice from 3 pears and 8 ounces of orange juice from 2 oranges. She makes a

pear-orange juice blend from an equal number of pears and oranges. What percent of the blend

is pear juice?

12. If x and y are non-zero numbers such that x is p% of y and y is 4p% of x, find p?

13. Evaluate

1002 − 99

2+ 98

2 − 972+ . . .+ 2

2 − 12.

14. A polygon has six times as many diagonals as it has sides. How many vertices does the polygon

have?

15. In Plano, there were 777 people who voted. There were 10% more female voters than male

voters. How many female voters were there?

16. (a) Find the greatest prime p such that p2 divides 95! + 96! + 97!

(b) Find the second greatest prime with this property.

17. What is the remainder when 1! + 2! + 3! + . . .+ 19! is divided by 100?

More Problems

18. The sum of 21 consecutive integers is 378. Find the least of these numbers.

19. What is the greatest multiple of 8 whose digits are all different?

20. N is a six-digit number and the sum of its digits is 37. The sum of the digits of N + 1 is 2.

Find N .

21. The ratio of Science and Arts students in a college is 4:3. If 14 Science students shift to Arts

then the ratio becomes 1:1. Find the total number of Science and Arts students.

22. The average age of five people in a room is 30 years. An 18-year-old person leaves the room.

What is the average age of the four remaining people?

23. Carol has three test scores of 84, 90, and 86. There is one more test during the semester. She

wants to make an A in the class, which means she needs her average to be a 90 or more. All

four tests count the same14 of the total grade. What would be the grade she needs to make on

the fourth test to make an A in the class?

24. What is the average of the first 50 positive integers?

25. Find the greatest n for which n! ends in exactly 33 zeros.

26. In the addition, where different letters represent different digits,

A

B

CD

EF

+ GH

XY

find X and Y.

27. Write 1,000,000 as a sum of a prime number and a perfect square.

28. The sum of fifteen consecutive integers is 105. Find their product.

29. At a reception with 25 participants every two people shake hands with one another. How many

handshakes occur?

30. If

1− 12010

1− 150

=m

n

where m and n are positive integers with no common divisor, then m− n equals

(A) 1 (B) 2 (C) 3 (D) 4 (E) 5

31. Jimmy took a three-digit number and reversed its digits. He then added the two numbers. The

sum was 1110. The middle digit of the original number is

(A) 1 (B) 2 (C) 3 (D) 4 (E) 5

32. If 53232

5is written in decimal form, the sum of its digits is

(A) 5 (B) 23 (C) 32 (D) 35 (E) 53

Solutions


Solution: Given equation can be rewritten as 10a+ b+ 10b+ c+ 10c+ a= 100a + 10b + c, from which 89a = 10c + b. Since a, b, c are digits, 10c + b < 100 so a = 1.

Then we see that b = 9 and c = 8. Therefore, the only solution is 198.




2. Evaluate:

a = (128− 2)(128− 22)(128− 2

3) . . . (128− 2

7)(128− 2

8).

Solution: Since 128 = 27, the last factor is 2

7 − 27= 0, hence the product is 0.




2. Evaluate:

a = (128− 2)(128− 22)(128− 2

3) . . . (128− 2

7)(128− 2

8).



3. Show that (3201

+ 3204

)÷ (3201 − 3

200+ 3


Solution: We can write

(3201

+ 3204

)÷ (3201 − 3

200+ 3

199) = 3

201 · (1 + 33)÷ [3

199(3

2 − 3 + 1)]

= 32 · (1 + 27)÷ (9− 3 + 1)

= 32 · 28÷ 7

= 32 · 22

= 62




2. Evaluate:

a = (128− 2)(128− 22)(128− 2

3) . . . (128− 2

7)(128− 2

8).



3. Show that (3201

+ 3204

)÷ (3201 − 3

200+ 3



(3201

+ 3204

)÷ (3201 − 3

200+ 3

199) = 3

201 · (1 + 33)÷ [3

199(3

2 − 3 + 1)]

= 32 · (1 + 27)÷ (9− 3 + 1)

= 32 · 28÷ 7

= 32 · 22

= 62



Solution: If our ages are x and y, we can write

x− y = 30

x

2= (y + 8)

From the first equation we can write y = x− 30. Substitution into the second equation yields

x

2= x− 30 + 8 therefore x = 44

Substituting that value into y = x− 30 we get y = 14.




2. Evaluate:

a = (128− 2)(128− 22)(128− 2

3) . . . (128− 2

7)(128− 2

8).



3. Show that (3201

+ 3204

)÷ (3201 − 3

200+ 3



(3201

+ 3204

)÷ (3201 − 3

200+ 3

199) = 3

201 · (1 + 33)÷ [3

199(3

2 − 3 + 1)]

= 32 · (1 + 27)÷ (9− 3 + 1)

= 32 · 28÷ 7

= 32 · 22

= 62




x− y = 30

x

2= (y + 8)


x

2= x− 30 + 8 therefore x = 44



Solution: Since 12 = 22 · 3, while 11 and 13 are prime, these numbers have no common factors

– therefore all of their common multiples are 11 · 12 · 13 · n, (n = 1, 2, . . .). The only 4-digit

multiples of 11 · 12 · 13 are 1716, 3432, 5148, 6864, and 8580. Among them only 5148 does not

have the desired form.




2. Evaluate:

a = (128− 2)(128− 22)(128− 2

3) . . . (128− 2

7)(128− 2

8).



3. Show that (3201

+ 3204

)÷ (3201 − 3

200+ 3



(3201

+ 3204

)÷ (3201 − 3

200+ 3

199) = 3

201 · (1 + 33)÷ [3

199(3

2 − 3 + 1)]

= 32 · (1 + 27)÷ (9− 3 + 1)

= 32 · 28÷ 7

= 32 · 22

= 62




x− y = 30

x

2= (y + 8)


x

2= x− 30 + 8 therefore x = 44



Solution: Since 12 = 22 · 3, while 11 and 13 are prime, these numbers have no common factors

– therefore all of their common multiples are 11 · 12 · 13 · n, (n = 1, 2, . . .). The only 4-digit

multiples of 11 · 12 · 13 are 1716, 3432, 5148, 6864, and 8580. Among them only 5148 does not

have the desired form.

6. Prove that

145678 + 456781 + 567814 + 678145 + 781456 + 814567

is the product of six different prime numbers.

Solution: Write 145678 = 1 · 105 + 4 · 104 + 5 · 103 + 6 · 102 + 7 · 10 + 8 and do the same for the

remaining numbers. Because each of the digits 1, 4, 5, 6, 7, 8 appears exactly once at each of the

six place value positions, the result of addition is

(1 + 4 + 5 + 6 + 7 + 8) · (105 + 104+ 10

3+ 10

2+ 10 + 1) = 31 · 111 · 1001

= 31 · 3 · 37 · 11 · 7 · 13

7. Find the last digit of 21+2+3+...+2009

.

Solution: We have 21+2+3+...+2009

= 22009·2010

2 = 22009·1005. The integer 2009 · 1005 is of the form

4s+ 1, hence the last digit of 22009·1005

is 2.

7. Find the last digit of 21+2+3+...+2009.

Solution: We have 21+2+3+...+2009 = 22009·2010

2 = 22009·1005. The integer 2009 · 1005 is of the form4s+ 1, hence the last digit of 22009·1005 is 2.


Solution: Since 50! has more factors of 2 than factors of 5, we need to find the exponent of 5 in50! = 1 · 2 · 3 · . . . · 50. This exponent is given by the factors

5, 2 · 5, 3 · 5, 4 · 5, 52, 6 · 5, 7 · 5, 8 · 5, 9 · 5, 2 · 52

hence it is 12. It follows that 50! ends by 12 zeros


Solution: We have 21+2+3+...+2009 = 22009·2010




5, 2 · 5, 3 · 5, 4 · 5, 52, 6 · 5, 7 · 5, 8 · 5, 9 · 5, 2 · 52


9. The average cost of a long-distance call in 1985 was 41 cents per minute, and the average costof a long-distance call in 2005 was 7 cents per minute. Find the approximate percent decreasein the cost per minute of a long-distance call.

Solution: The difference in the cost is 41−7 = 34 cents. The percent decrease is 100 · 3441 ≈ 83%.


Solution: We have 21+2+3+...+2009 = 22009·2010




5, 2 · 5, 3 · 5, 4 · 5, 52, 6 · 5, 7 · 5, 8 · 5, 9 · 5, 2 · 52




10. An athlete’s target heart rate, in beats per minute, is 80% of the theoretical maximum heartrate. The maximum heart rate is found by subtracting the athlete’s age, in years, from 220. Tothe nearest whole number, what is the target heart rate of an athlete who is 26 years old?

Solution: A 26-year-old’s target heart rate is 0.8 · (220 − 26) = 155.2 beats per minute. Thenearest whole number is 155.


Solution: We have 21+2+3+...+2009 = 22009·2010




5, 2 · 5, 3 · 5, 4 · 5, 52, 6 · 5, 7 · 5, 8 · 5, 9 · 5, 2 · 52






11. Miki has a dozen oranges of the same size and a dozen pears of the same size. Miki can extract8 ounces of pear juice from 3 pears and 8 ounces of orange juice from 2 oranges. She makes apear-orange juice blend from an equal number of pears and oranges. What percent of the blendis pear juice?

Solution: For convenience, let us use 6 pears to make 16 oz of pear juice and 6 oranges to make24 oz of orange juice for a total of 40 oz of juice. The percent of the pear juice is 16

40 = 410 = 40%


Solution: We have 21+2+3+...+2009 = 22009·2010




5, 2 · 5, 3 · 5, 4 · 5, 52, 6 · 5, 7 · 5, 8 · 5, 9 · 5, 2 · 52








40 = 410 = 40%


Solution: We have x = py100 and y = 4px

100 .

Multiplying the two equalities and then dividing by xy yields 1 = 4p2

104 .It follows that p = 50.


Solution: We have 21+2+3+...+2009 = 22009·2010




5, 2 · 5, 3 · 5, 4 · 5, 52, 6 · 5, 7 · 5, 8 · 5, 9 · 5, 2 · 52








40 = 410 = 40%



100 .



13. Evaluate1002 − 992 + 982 − 972 + . . .+ 22 − 12.

Solution: Use a2 − b

2 = (a+ b)(a− b) to obtain

1002 − 992 + 982 − 972 + . . .+ 22 − 12 = (100 + 99)(100− 99) + (98 + 97)(98− 97) + . . .+ (2 + 1)(2− 1)

= 100 + 99 + . . .+ 1 =100 · 101

2= 5050


Solution: We have 21+2+3+...+2009 = 22009·2010




5, 2 · 5, 3 · 5, 4 · 5, 52, 6 · 5, 7 · 5, 8 · 5, 9 · 5, 2 · 52








40 = 410 = 40%



100 .



13. Evaluate1002 − 992 + 982 − 972 + . . .+ 22 − 12.

Solution: Use a2 − b

2 = (a+ b)(a− b) to obtain

1002 − 992 + 982 − 972 + . . .+ 22 − 12 = (100 + 99)(100− 99) + (98 + 97)(98− 97) + . . .+ (2 + 1)(2− 1)

= 100 + 99 + . . .+ 1 =100 · 101

2= 5050

14. A polygon has six times as many diagonals as it has sides. How many vertices does the polygonhave?

Solution: Each vertex of a polygon is connected to n−1 vertices, two of which are not diagonals,but sides, so each vertex is connected to n − 3 diagonals. The product n(n − 3) counts eachdiagonal twice, so there are a total of n(n−3)

2 diagonals. From the problem statement we know

that n(n−3)2 = 6n. Since n �= 0, we can divide both sides by n and multiply both sides by 2 to

get n− 3 = 12, i.e., n = 15.

15. In Plano, there were 777 people who voted. There were 10% more female voters than malevoters. How many female voters were there?

Solution: Let x be the number of male voters. Then there are 11x10 female voters. We know that

x+11x

10=

21x

10= 777

hence x = 370. In conclusion there are 777− 370 = 407 female voters.



x+11x

10=

21x

10= 777




Solution: (a) Note that

95! + 96! + 97! = 95!(1 + 96 + 96 · 97) = 95!(97 + 96 · 97)= 95! · 97(1 + 96) = 95! · 972

We know that 97 is a prime number and it is clear that 972 does divide the expression we aregiven. Any prime larger than 97 will simply not divide the given expression.

(b) We find that 47 divides 47 and 94 hence 472 | 95! ·972. Any prime larger than 47 and smallerthan 97 will divide 95! · 972 but its square will not.



x+11x

10=

21x

10= 777





95! + 96! + 97! = 95!(1 + 96 + 96 · 97) = 95!(97 + 96 · 97)= 95! · 97(1 + 96) = 95! · 972




Solution: Any of the numbers 10!, 11!, . . . , 19! are divisible by 100. Hence we can reduce theproblem to finding the remainder of 1!+2!+ . . .+9! divided by 100. Notice that 8!+9! = 8! ·10which is also divisible by 100. Finally,

1! + 2! + . . .+ 7! = 1 + 2 + 6 + 24 + 120 + 720 + 5040 = 5913

Hence the remainder is 13.



x+11x

10=

21x

10= 777





95! + 96! + 97! = 95!(1 + 96 + 96 · 97) = 95!(97 + 96 · 97)= 95! · 97(1 + 96) = 95! · 972





1! + 2! + . . .+ 7! = 1 + 2 + 6 + 24 + 120 + 720 + 5040 = 5913



Solution: Setting up an equation like above we find 21x + 20·212 = 378 so the least number is

x = 8.



x+11x

10=

21x

10= 777





95! + 96! + 97! = 95!(1 + 96 + 96 · 97) = 95!(97 + 96 · 97)= 95! · 97(1 + 96) = 95! · 972





1! + 2! + . . .+ 7! = 1 + 2 + 6 + 24 + 120 + 720 + 5040 = 5913




x = 8.


Solution: Clearly, we must use all ten digits in order to get the greatest multiple a1a2 . . . a10.Notice that a1a2 . . . a7000 is divisible by 8. In order to get the maximum, we assign a1a2 . . . a7 =9876543. The last three digits are 2, 1, 0 and they must form a number that is divisible by 8.Observe that a8 �= 2, because neither 210 nor 201 is divisible by 8. Hence a8 = 1 and a9 = 2,a10 = 0, yielding 9876543120 as the greatest multiple of 8 whose digits are all distinct.



x+11x

10=

21x

10= 777





95! + 96! + 97! = 95!(1 + 96 + 96 · 97) = 95!(97 + 96 · 97)= 95! · 97(1 + 96) = 95! · 972





1! + 2! + . . .+ 7! = 1 + 2 + 6 + 24 + 120 + 720 + 5040 = 5913




x = 8.


Solution: Clearly, we must use all ten digits in order to get the greatest multiple a1a2 . . . a10.Notice that a1a2 . . . a7000 is divisible by 8. In order to get the maximum, we assign a1a2 . . . a7 =9876543. The last three digits are 2, 1, 0 and they must form a number that is divisible by 8.Observe that a8 �= 2, because neither 210 nor 201 is divisible by 8. Hence a8 = 1 and a9 = 2,a10 = 0, yielding 9876543120 as the greatest multiple of 8 whose digits are all distinct.

20. N is a six-digit number and the sum of its digits is 37. The sum of the digits of N + 1 is 2.Find N .

Solution: The last digit must be 9 otherwise N +1 cannot have the sum of its digits equal to 2.In the same manner we can show that the last 4 digits of N are all 9. Thus we conclude thatN = 109999.

21. The ratio of Science and Arts students in a college is 4:3. If 14 Science students shift to Artsthen the ratio becomes 1:1. Find the total number of Science and Arts students.

Solution: Let x be the number of Science students and let y be the number of Arts students.Then from the problem we know two things:

x

y=

4

3and

x− 14

y + 14= 1



x

y=

4

3and

x− 14

y + 14= 1

22. The average age of five people in a room is 30 years. An 18-year-old person leaves the room.What is the average age of the four remaining people?

Solution: Originally, the sum of the ages of the people in the room is 5 · 30 = 150. After the18-year-old leaves, the sum of the ages of the remaining people is 150−18 = 132. So the averageage of the four remaining people is 132

4 = 33 years.



x

y=

4

3and

x− 14

y + 14= 1



4 = 33 years.

23. Carol has three test scores of 84, 90, and 86. There is one more test during the semester. Shewants to make an A in the class, which means she needs her average to be a 90 or more. Allfour tests count the same 1

4 of the total grade. What would be the grade she needs to make onthe fourth test to make an A in the class?

Solution: An average of 90 will guarantee Carol an A in the class.To average a 90 on all 4 tests she needs to have accumulated a total of 4 · 90 = 360 points inthe four tests. Right now she has 84 + 90 + 86 = 260 points. Hence to have a chance at an Ain the class she needs to score 360− 260 = 100 on the last test.



x

y=

4

3and

x− 14

y + 14= 1



4 = 33 years.





Solution: The sum of the first 50 positive integers is

1 + 2 + 3 + . . .+ 50 =50 · 51

2= 1275

Hence their average is 127550 = 25.5



x

y=

4

3and

x− 14

y + 14= 1



4 = 33 years.






1 + 2 + 3 + . . .+ 50 =50 · 51

2= 1275



Solution: We use the same idea as in exercise 4: the number of zeros at the end of factorialis the number of 5’s in its product. A simple calculation shows that 125! has 25 + 5 + 1 zerosat the end because there are 25 numbers divisible by 5, five numbers divisible by 52 and onenumber that is divisible by 53. So 125! has 31 zeros at the end. It follows that 130! has 32zeros, 135! has 33 zeros, and 140! has 34 zeros. Thus the greatest integer for which n! ends inexactly 33 zeros is n = 139.



x

y=

4

3and

x− 14

y + 14= 1



4 = 33 years.






1 + 2 + 3 + . . .+ 50 =50 · 51

2= 1275



Solution: We use the same idea as in exercise 4: the number of zeros at the end of factorialis the number of 5’s in its product. A simple calculation shows that 125! has 25 + 5 + 1 zerosat the end because there are 25 numbers divisible by 5, five numbers divisible by 52 and onenumber that is divisible by 53. So 125! has 31 zeros at the end. It follows that 130! has 32zeros, 135! has 33 zeros, and 140! has 34 zeros. Thus the greatest integer for which n! ends inexactly 33 zeros is n = 139.

26. In the additionA

B

CD

EF

+ GH

XY

find X and Y.

Solution: BecauseA+B + . . .+H +X + Y = 0 + 1 + . . .+ 9 = 45

when 45 − (X + Y ) is divided by 9, the remainder is equal to the remainder when X + Y isdivided by 9. This means that X + Y is divisible by 9, and since X and Y are different digits,X + Y < 18, so X + Y = 9. Since A+ B + CD + EF +GH ≥ 4 + 5 + 10 + 26 + 37 = 82, wemust have XY = 90. It follows that X = 9 and Y = 0.


Solution: Let p be a prime number and x and integer number such that p + x2 = 1, 000, 000

then p = (1000− x)(1000 + x). This implies that 1000− x = 1 and 1000 + x = p which yieldsp = 1999, which is indeed a prime. Then x = 999, so x

2 = 998001.

when 45− (X+Y ) is divided by 9, the remainder is equal to the remainder when X+Y is dividedby 9. This means that X + Y is divisible by 9, and since X and Y are different digits, X + Y < 18,so X + Y = 9. Since A+B +CD+EF +GH ≥ 4+ 5+ 10+ 26+ 37 = 82, we must have XY = 90.It follows that X = 9 and Y = 0.




2 = 998001.


Solution: Note that the sum of the first fifteen consecutive numbers is

1 + 2 + . . .+ 15 = 120

Thus one of the numbers in our sum is 0. We have 0+1+ . . .+14 = 105 then 0 · 1 · . . . · 14 = 0,and we are done





2 = 998001.



1 + 2 + . . .+ 15 = 120


29. At a reception with 25 participants every two people shake hands with one another. How manyhandshakes occur?

Solution: Each of 25 participant shakes hands with 24 other participants. If we multiply 25 ·24,we are counting each handshake twice, so the solution is

25 · 242

= T24





2 = 998001.



1 + 2 + . . .+ 15 = 120


29. At a reception with 25 participants every two people shake hands with one another. How manyhandshakes occur?

Solution: Each of 25 participant shakes hands with 24 other participants. If we multiply 25 ·24,we are counting each handshake twice, so the solution is

25 · 242

= T24

30. If1− 1

2010

1− 150

=m

n

where m and n are positive integers with no common divisor, then m− n equals

(A) 1 (B) 2 (C) 3 (D) 4 (E) 5

Solution: The answer is (D). We have

m

n=

200920104950

=2009

2010· 5049

=49 · 41201 · 10 · 5 · 10

49=

205

201

where 205 and 201 are relatively prime. The answer is

205− 201 = 4

31. Jimmy took a three-digit number and reversed its digits. He then added the two numbers. Thesum was 1110. The middle digit of the original number is

(A) 1 (B) 2 (C) 3 (D) 4 (E) 5

Solution: The answer is (E). If the original number was abc, then

1110 = abc+ cba = (100a+ 10b+ c) + (100c+ 10b+ a)

= 101(a+ c) + 20b

Then 101(a+ c) = 1110− 20b is divisible by 10, so a+ c is divisible by 10. Because a and c aredigits, 0 < a+ c < 20, so a+ c = 10. It follows that 1010 = 1110− 20b, so b = 5.

31. Jimmy took a three-digit number and reversed its digits. He then added the two numbers. Thesum was 1110. The middle digit of the original number is

(A) 1 (B) 2 (C) 3 (D) 4 (E) 5

Solution: The answer is (E). If the original number was abc, then

1110 = abc+ cba = (100a+ 10b+ c) + (100c+ 10b+ a)

= 101(a+ c) + 20b

Then 101(a+ c) = 1110− 20b is divisible by 10, so a+ c is divisible by 10. Because a and c aredigits, 0 < a+ c < 20, so a+ c = 10. It follows that 1010 = 1110− 20b, so b = 5.

32. If 532325 is written in decimal form, the sum of its digits is

(A) 5 (B) 23 (C) 32 (D) 35 (E) 53

Solution: The answer is (B). Because 325 = (25)5 = 225 and 532 = 57525, the given number canbe written as

57 · (5 · 2)25 = 78125 · 1025 = 7812500 . . . 0 (25 zeros)

The sum of the digits for the decimal representation is

7 + 8 + 1 + 2 + 5 = 23

amc 8 at mmc - metroplex math circle · 8 years older than she really is, we would be the same age....

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