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MOCK EXAMINATION

AMC 12 American Mathematics Contest 12

Test Sample

Detailed Solutions

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It’s one of the best ways to get ready for the AMC.

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AMC 10 Mock Test

Detailed Solutions

Problem 1

Answer: (E)

Solution 1

Note that there is more than 1 four-legged table. So there are at least 2 four-legged tables.

Since there are 23 legs in total, there must be fewer than 6 four-legged tables, which have × = legs. Thus, there are between 2 and 5 four-legged tables.

If there are 2 four-legged tables, then these tables account for × = legs, leaving − = legs for the three-legged tables, which implies that there are 5 three-legged tables.

(We can check that if there are 3 or 4 four-legged tables, then the number of remaining legs is

not divisible by 3, and if there are 5 four-legged tables, then there is only 1 three-legged table,

which is not allowed.)

Solution 2

Since there is more than 1 table of each type, then there are at least 2 three-legged tables and

2 four-legged tables. These tables account for + = legs.

There are − = more legs that need to be accounted for. These must come from a

combination of three-legged and four-legged tables. The only way to make 9 from 3s and 4s is to

use three 3s.

Therefore, there are + = three-legged tables and 2 four-legged tables.

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Problem 2

Answer: (B)

Note that = + − − = − = . So = .

Problem 3

Answer: (E)

The darkened curve intersects line l in three points. Let be the point of intersection that lies

on , as shown in the figure below.

Then is a radius of the semicircle on the left, is a radius of the semicircle on the right,

and + = = . Since the darkened curve consists of the two semicircles, its length is the sum of the lengths of

the two semicircles: 𝜋 ∙ + 𝜋 ∙ = 𝜋 + = 𝜋.

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Problem 4

Answer: (D)

By the given conditions, + + = , + = .

Thus, = − = −

Problem 5

Answer: (A)

If , , and are the roots of − + = , then by Vieta's theorem, + + = , + + = − , and = − . Thus, + + + = + + + + + + += − + − + + = − .

Problem 6

Answer: (D)

Brad could have traveled at most × =

miles.

Each 5-digit palindrome is uniquely determined by its first three digits. The next palindromes

after 38983 are

39093, 39193, 39293, and 39393.

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Note that − = > . So this number and all larger ones are too large.

Since − = > , the difference is too far to drive in 5 hours without exceeding the speed limit of 70 miles per

hour. Brad could have driven − =

miles during the 5 hours for an average speed of =

miles per hour.

Problem 7

Answer: (D)

Since = + is the inverse function of = √ −7 , we have: ∘ = − ∘ − = . Thus, − ∘ − = .

Problem 8

Answer: (C)

Since Alex needs 12 hours to shovel all of his snow, he shovels of his snow per hour.

Since Bob needs 8 hours to shovel all of Alex's snow, he shovels of Alex's snow per hour.

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Similarly, Carl shovels of Alex's snow every hour, and Dick shovels of Alex's snow per hour.

Together, Alex, Bob, Carl, and Dick can shovel + + + =

of Alex's snow per hour.

Therefore, together they can shovel

=

of Alex's snow per minute.

Thus, by shoveling of Alex's snow per minute, together they will shovel all of Alex's snow

in 96 minutes.

Problem 9

Answer: (C)

Squaring the both sides of the equation sin + cos = gives: sin + ∙ ∙ sin cos + cos = . Using sin + cos = , we have: − cos + ∙ ∙ sin cos + − sin = , The equation can be simplified to sin − ∙ ∙ sin cos + cos = , which is equivalent to sin − cos = . It follows that sin = cos

which implies that

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tan = .

Problem 10

Answer: (A)

Together, Hose X and Hose Y fill the pool in 10 hours.

Thus, it must take Hose X more than 6 hours to fill the pool when used by itself.

Therefore, , since is a positive integer.

Similarly, it must take Hose Y more than 10 hours to fill the pool when used by itself.

Therefore, , since is a positive integer.

When used by itself, the fraction of the pool that Hose X fills in 10 hours is .

When used by itself, the fraction of the pool that Hose Y fills in 10 hours is .

When used together, Hose X and Hose Y fill the pool once in 10 hours. Thus, + = , which is equivalent to − − = . Completing the rectangle gives: − − = . Note that the prime factorization of 100 is = ∙ . Thus, 100 has + + =

distinct factors.

Correspondingly, − , − has 9 positive solutions. Hence, there are only 9 different

possible values for . In fact, 9 positive solutions are:

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− , − = , , , , , , , , , , , , , , , , , . That is, , = , , , , , , , , , , , , , , , , , .

Problem 11

Answer: (D)

From the equation − + − + = , we have 3 cases:

− + = , or − + = , or − + = − and − + is even.

Case 1: − + = . We get: = or = .

Case 2: − + = . We get: = or = .

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Case 3: − + = − and − + is even.

The following 2 solutions satisfy both conditions: = or = . Therefore, the equation has 5 integer solutions.

Problem 12

Answer: (B)

Plugging in = into + = gives: + =

for every real number . Note that = . So for any integer 𝑛 > , we have: 𝑛 = 𝑛 − = 𝑛 − = ⋯ = = = . Hence, = .

Problem 13

Answer: (B)

Note that log + log + log + log = log + log + log + log

= log ∙ ∙ ∙ .

Thus, log ∙ ∙ ∙ = . Raising both sides to the tenth power gives: ∙ ∙ ∙ = .

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So = = , = = . Hence, + + + = .

Problem 14

Answer: (D)

Solution 1

Label the vertices of the new figure as shown below.

When the paper is folded in this way, the portion of the original bottom face of the paper that is

visible has the same area as the original portion of the top side of the paper to the right of the

fold. This is quadrilateral . Of the portion of the original sheet to the left of the fold, the part, which is hidden and thus not

included in the area of the new figure, is the triangular portion under the folded part. This is the

section under ∆ , which is an isosceles right triangle with side lengths of 8. The hidden

triangle is congruent to ∆ . Thus, the area of the portion of the original top face of the paper that is visible is the area to the

left of the fold, minus the area of the hidden triangle.

Therefore, the area of the new figure equals the area of the original rectangle minus the area of ∆ :

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× − × = .

Solution 2

Let = . Since + + = , we have = − − = − − = − . Note that ∆ is a right triangle with base and height both equal to 8. So the area of the new

figure is equal to: Area Rectangle + Area ∆ + Area Rectangle = + × + − = .

Problem 15

Answer: (A)

Solution 1

Note that ( + 𝑖√ ) = + 𝑖 + 𝑖 + 𝑖 + 𝑖 = −

and

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( − 𝑖√ ) = − 𝑖 + 𝑖 − 𝑖 + 𝑖 = − . Since = × + , and = × + , we have

+ = − ∙ + 𝑖√ + − ∙ − 𝑖√ + − ∙ + 𝑖√ + − ∙ − 𝑖√

= .

Solution 2

Note that

√ = sin 𝜋 = cos 𝜋. Thus, 𝑘 = cos 𝜋 + 𝑖 sin 𝜋 𝑘 + cos 𝜋 − 𝑖 sin 𝜋 𝑘

= cos 𝜋 + 𝑖 sin 𝜋 𝑘 + cos − 𝜋 + 𝑖 sin − 𝜋 𝑘 . Using De Moivre's formula, we have: 𝑘 = (cos 𝑘𝜋 + 𝑖 sin 𝑘𝜋) + (cos (− 𝑘𝜋) + 𝑖 sin (− 𝑘𝜋))

= cos 𝑘𝜋. Hence, + = cos 𝜋 + cos 𝜋

= cos ( ∙ 𝜋 + 𝜋) + cos ∙ 𝜋 + 𝜋

= cos ( 𝜋) + cos 𝜋 = .

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Problem 16

Answer: (B)

Using the sum formula for tangent, we have: tan + = tan + tan− tan tan . Note that tan + = tan = . Thus, tan + tan− tan tan = , which implies that − tan tan = tan + tan . Then + tan + tan = + tan + tan + tan tan= + − tan tan + tan tan = .

Problem 17

Answer: (D)

Denote by the first point that is picked. Let and be the points on the circle which are

exactly 1 unit away from . Then = = = = = , where denotes the center of the circle.

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The triangles and are equilateral and the arc has angle . Hence, two-thirds of

the points on the circle are at least 1 unit away from . Therefore, the probability that the chord

joining the two points has length at least 1 is .

Problem 18

Answer: (D)

Solution 1

Let the millipede try to put on all 𝑛 things in a random order. Each of the ( 𝑛 ! permutations is

equally probable. For any fixed leg, the probability that the millipede first puts on the sock and

only then the shoe is clearly . Then the probability that the millipede correctly puts things on all

legs is 𝑛. Therefore, the number of correct permutations must be 𝑛 !𝑛 .

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Solution 2

Number the millipede's legs from 1 through 𝑛, and let 𝑘 and 𝑘 denote the sock and shoe that

will go on leg k.

A possible arrangement of the socks and shoes is a permutation of the 𝑛 symbols , , ⋯ , 𝑛, 𝑛 in which 𝑘 precedes 𝑘 for 𝑘 𝑛.

There are (2 𝑛 ! permutations of the 𝑛 symbols, and precedes in exactly half of these, or 𝑛 !

permutations.

Similarly, precedes in exactly half of those, or 𝑛 !

permutations. Continuing, we can conclude that 𝑘 precedes 𝑘 for 𝑘 𝑛 in exactly 𝑛 !

permutations.

Problem 19

Answer: (B)

Since is rectangular, then ∠ = ∠ = . Also, = = and = = . Note that ∆ and ∆ are two congruent 3-4-5 right triangles with legs of 15 and 20. So the

hypotenuse length is = . Draw perpendiculars from and to and , respectively, on . Also, join to .

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By the AA similarity postulate, right ∆ is similar to right ∆ because they share ∠ . So ∆ is also a 3-4-5 right triangle with hypotenuse of 15, and thus, ∶ ∶ = ∶ ∶ , which implies that = , = . Similarly, we have: = , = . Thus, = − − = − − = . Using the 3-D Pythagorean Theorem, we obtain: = √ + + = √ + + = √ .

Problem 20

Answer: (E)

Solution 1

Alan gives 24 bars that account for 45% of the total weight to Bob. Thus, each of these 24 bars

accounts for an average of

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% = %

of the total weight.

Alan gives 13 bars that account for 26% of the total weight to Carl. Thus, each of these 13 bars

accounts for an average of % = %


Note that the bars given to Dale account for % − % − % = % of the total weight.

Let be the number of bars that Dale received. Then each of these bars accounts for an

average of % = %


Since each of the bars that she gives to Dale is heavier than each of the bars given to Bob (which

were the 24 lightest bars) and is lighter than each of the bars given to Carl (which were the 13

heaviest bars), then the average weight of the bars given to Dale must be larger than % and

smaller than 2%. Thus, % % %, or = =

Hence, there is only one solution: = .

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Solution 2

Without loss of generality, assume that the total weight of all of the bars is 100. Then the bars

given to Bob weigh 45 and the bars given to Carl weigh 26.

Suppose that Alan gives 𝑛 bars to Dale. These bars weigh − − = . Let < < ⋯ <

be the weights of the 24 bars given to Bob.

Let < < ⋯ <

be the weights of the 13 bars given to Carl.

Let < < ⋯ < 𝑛

be the weights of the 𝑛 bars given to Dale.

Note that < < ⋯ < < < < ⋯ < < < < ⋯ < 𝑛

since the lightest bars are given to Bob and the heaviest to Carl.

Also, + + ⋯ + = , + + ⋯ + = , + + ⋯ + 𝑛 = . Now is the heaviest of the bars given to Bob, so = + + ⋯ + <

and so > = . Also, is the lightest of the bars given to Carl, so = + + ⋯ + 𝑛 > . and so

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< . But each of the 𝑛 bars given to Dale is heavier than and each is lighter than . Thus, 𝑛 < + + ⋯ + < 𝑛 , or 𝑛 < < 𝑛 . It follows that 𝑛 < 𝑛 < < 𝑛 < 𝑛, and so 𝑛 < ∙ = = , and 𝑛 > = . Note that 𝑛 is an integer. So we have only one solution: 𝑛 = . So Dale receives 15 bars.

Problem 21

Answer: (D)

The maximum is obtained when no three planes pass through the same diameter of the sphere.

Let denote the surface of the sphere. One plane divides into two regions. The second plane

intersects the first plane in 2 points on , thus adding two more regions. The third plane

intersects each of the previous planes on S at two points which means we will get ×

additional regions. The 𝑛th plane intersects the previous 𝑛 − planes at 𝑛 − points, which

means it would add 𝑛 − regions on . Hence, the total number of regions is + + × + × + ⋯ + × = + × × = .

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Problem 22

Answer: (A)

Let be the probability that the sum will reach ± if it is presently . If the sum is zero to

start, then there is a probability of ∙

that it will be 1 after one trial, and ∙

that it will be − . By symmetry, = − . Expanding probabilities gives = ∙ + ∙ − = ,

= + , = + .

Solving the about 3 equations simultaneously gives = . Thus, the probability that the sum never reaches 3 is: − = .

Problem 23

Answer: (D)

Note that the grid is a 5 by 5 grid of squares and each square has side length 10 units, then the

whole grid is 50 by 50.

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Since the diameter of the coin is 8 units, the radius of the coin is 4 units.

We consider where the center of the coin lands when the coin is tossed, since the location of the

center determines the position of the coin.

Since the coin lands so that no part of it is off of the grid, then the center of the coin must land at

least 4 units away from each of the outer edges of the grid. This means that the center of the coin

lands anywhere in the region extending from 4 units from the left edge to 4 units from the right

edge (a length of − − = units) and from 4 units from the top edge to 4 units to the

bottom edge (a width of − − = units). Thus, the center of the coin must land in a

square that is 42 by 42 in order to land so that no part of the coin is off the grid.

Therefore, the total admissible area in which the center can land is × . Consider one of the 25 squares. For the coin to lie completely inside the square, its center must

land at least 4 units from each edge of the square.

As illustrated above, it must land in a region of length − − = and of width − − = . There are 25 possible such regions (one for each square) so the area in which the center of the

coin can land to create a winning position is × × . Therefore, the probability that the coin lands in a winning position is equal to the area of the

region in which the center lands giving a winning position, divided by the area of the region in

which the coin may land, or × ×× = × = .

Problem 24

Answer: (A)

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Squaring the both sides of the equation = + + gives: = + + = + + + + + = + + + , which implies that + + = . Thus, + + = + + . Rearranging and multiplying both sides by 2, we have: + + − + + = . Completing the squares yields

− + − + − = . Hence, = = , which implies that the expression has only one possible value: + + = . The number of possible values of the expression is 1.

Problem 25

Answer: (E)

Let 𝑢 𝑛 denote the units digit of the positive integer 𝑛. We make three important notes:

It is the final position on the circle for which we are looking, not the total number of

times travelled around the circle. Therefore, moving 63 steps, for example, around the

circle is equivalent to moving 3 steps around the circle because in both cases the counter

ends up in the same position. Since 10 steps gives one complete trip around the circle,

then we only care about the units digit of the number of steps. That is, 𝑢 𝑛𝑛 .

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To determine the final position, we want to find the sum of the number of steps for each

of the 1234 moves; that is, we want to determine = + + ⋯ + + . In fact, we only need to calculate the units digit of the sum of the numbers of steps 𝑢 , which is equal to 𝑢(𝑢 + 𝑢 + ⋯ + 𝑢 + 𝑢 ).

To calculate 𝑢 𝑛𝑛 , we only need to detect the units digit of the base 𝑛, 𝑢 𝑛 . Note that 𝑢 𝑛𝑛 = 𝑢((𝑢 𝑛 )𝑛). To actually perform this calculation, we can always truncate to the units digit at each step

because only the units digits affect the units digits.

After that, we consider the different possible values of 𝑢 𝑛 and determine a pattern of the units

digits of powers of 𝑛: If 𝑢 𝑛 is 0, 1, 5, or 6, then 𝑢 𝑛𝑘 is 0, 1, 5, or 6, respectively, for any positive integer 𝑘. If 𝑢 𝑛 = , then the units digits of powers of 𝑛 alternate 4, 6, 4, 6, and so on.

If 𝑢 𝑛 = , then the units digits of powers of 𝑛 alternate 9, 1, 9, 1, and so on.

If 𝑢 𝑛 = , then the units digits of powers of 𝑛 cycle as 2, 4, 8, 6, 2, 4, 8, 6, and so on.




Next, we determine 𝑢 𝑛𝑛 , based on u(n):

If u(n) is 0, 1, 5, or 6, then 𝑢 𝑛𝑛 is 0, 1, 5, or 6, respectively.

If 𝑢 𝑛 = , then 𝑢 𝑛𝑛 = , since the exponent is even so the units digit will be that

occurring in even positions in the pattern.

If 𝑢 𝑛 = , then 𝑢 𝑛𝑛 = , since the exponent is odd so the units digit will be that

occurring in odd positions in the pattern.

If 𝑢 𝑛 = , then 𝑢 𝑛𝑛 will be either 4 or 6, depending on the exponent n, because

the exponent is certainly even, but the pattern of units digits cycles with length 4

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If 𝑢 𝑛 = , then 𝑢 𝑛𝑛 will be either 4 or 6, depending on the exponent 𝑛, because

the exponent is certainly even, but the pattern of units digits cycles with length 4.

Since the units digits of the base n repeat in a cycle of length 10 and the units digits of the

powers of 𝑛 for a fixed 𝑛 repeat every 1, 2, or 4 powers, then 𝑢 𝑛𝑛 repeats in a cycle of

length 20 (the least common multiple of 10, 1, 2, and 4).

So 𝑢(𝑢 + 𝑢 + ⋯ + 𝑢 + 𝑢 )= 𝑢 + + + + + + + + + + + + + + ++ + + + = 𝑢 = . To calculate the total for 1234 steps, we note that 61 cycles of 20 bring us to 1220 steps. After

1220 steps, the units digit of the sum is 𝑢 ∙ = 𝑢 = . We then add the units digit of the sum of 14 more steps, starting at the beginning of the cycle, to

obtain a final position equal to the units digit of 𝑢 + + + + + + + + + + + + + + = 𝑢 = . Therefore, the final position is 7.

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