2013 AIME I Problems/Problem 11

Problem 11

Ms. Math's kindergarten class has 16 registered students. The classroom has a very large number, N, of play blocks which satisfies the conditions:

(a) If 16, 15, or 14 students are present in the class, then in each case all the blocks can be distributed in equal numbers to each student, and

(b) There are three integers $0 < x < y < z < 14$ such that when $x$ , $y$ , or $z$ students are present and the blocks are distributed in equal numbers to each student, there are exactly three blocks left over.

Find the sum of the distinct prime divisors of the least possible value of N satisfying the above conditions.

Solution

N must be some multiple of the LCM of 14, 15, and 16 = $2^{4} \cdot 3 \cdot 5 \cdot 7$ ; this LCM is hereby denoted $k$ and $N = nk$ .

1, 2, 3, 4, 5, 6, 7, 8, 10, and 12 all divide $k$ , so $x, y, z = 9, 11, 13$

We have the following three modulo equations:

$nk\equiv 3 \pmod{9}$

$nk\equiv 3 \pmod{11}$

$nk\equiv 3 \pmod{13}$

To solve the equations, you can notice the answer must be of the form 9*11*13* $x$ + 3 where x is an integer. This must be divisible by the LCM(14, 14, 15), which is 560*3. Therefore, (9*11*13 $x$ +3)/(560*3) = y, which is an integer. Factor out 3 and divide to get (429x+1)/(560) = y. Therefore, 429x+1=560y or 429x-560y=1. We can use Bezout's Identity to solve for the least of $x$ and $y$ . We find that the least $x$ is 171 and the least $y$ is 171. Plug it into 9*11*13x+3 and factor to get that the distinct prime divisors are 2,3,4, and 131.

$2 + 3 + 5 + 7 + 131 = \boxed{148}$ .

2013 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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2013 AIME I Problems/Problem 11

Problem 11

Solution

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