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 1

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 $N = qk$ .

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 $9\cdot 11\cdot 13\cdot m + 3$ where $m$ is an integer.

This must be divisible by LCM $(14, 15, 16)$ , which is $560\cdot 3$ .

Therefore, $(9\cdot 11\cdot 13m + 3)/(560\cdot 3) = q$ , which is an integer. Factor out 3 and divide to get $(429m+1)/(560) = q$ . Therefore, $429m+1=560q$ . We can use Bezout's Identity or a Euclidean Algorithm bash to solve for the least of $m$ and $q$ .

We find that the least $m$ is $171$ and the least $q$ is $131$ .

Since we want to factor $1680 \cdot q$ , don't multiply; we already know that the prime factors of $1680$ are $2$ , $3$ , $5$ , and $7$ , and since $131$ is prime, we have $2 + 3 + 5 + 7 + 131 = \boxed{148}$ .

Solution 2

Note that the number of play blocks is a multiple of the LCM of 16, 15, and 14. The value of this can be found to be $(16)(15)(7) = 1680$ . This number is also divisible by 1, 2, 3, 4, 5, 6, 7, 8, 10, and 12, thus, the three numbers $x, y, z$ are $9, 11, 13$ .

Thus, $1680k \equiv 3$ when taken mod 9, 11, 13. Since $1680$ is congruent to 6 mod 9 and 3 mod 13, and congruent to 8 mod 11, the number $k$ must be a number that is congruent to $1$ mod $13$ , $2$ mod $3$ (because $6$ is a multiple of $3$ , which is a factor of $9$ that can be divided out) and cause $8$ to become $3$ when multiplied under modulo 11.*

Looking at the last condition shows that $k \equiv 10$ mod 11 (after a bit of bashing) and is congruent to $1$ mod $13$ and $2$ mod $3$ as previously noted. Listing out the numbers congruent to 10 mod 11 and 1 mod 13 yield the following lists:

10 mod 11: 21, 32, 43, 54, 65, 76, 87, 98, 109, 120, 131...

1 mod 13: 14, 27, 40, 53, 66, 79, 92, 105, 118, 131, 144, 157, 170...

Both lists contain $x$ elements where $x$ is the modulo being taken, thus, there must be a solution in these lists as adding $11(13)$ to this solution yields the next smallest solution. In this case, $131$ is the solution for $k$ and thus the answer is $1680(131)$ . Since $131$ is prime, the sum of the prime factors is $2 + 3 + 5 + 7 + 131 = \boxed{148}$ .

$\text*$ A

Modulus Solution

It is obvious that $N=a*2^4*3*5*7$ and so the only $(mod 3)$ number of students are $9, 11, 13$ . Therefore, $N=1287*k+3$ . Try some approaches and you will see that this one is one of the few successful ones:

Start by setting the two $N$ equations together, then we get $1680a=1287k+3$ . Divide by 3. Note that since the RHS is $1 (mod 3)$ , and since $560$ is $2 (mod 3)$ , then $a=3b+2$ , where $b$ is some nonnegative integer, because $a$ must be $2 (mod 3)$ .

This reduces to $560 * 3b + 1119 = 429k$ . Now, take out the 11! With the same procedure, $b=11c-1$ , where $c$ is some nonnegative integer.

You also get $c=13d+4$ , at which point $k=171+560d$ . D CAN BE ZERO!! Therefore, $c=4, b=43, a=131$ , and we know the prime factors of $N$ are $2, 3, 5, 7, 131$ so the answer is $\boxed{148}$ !

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

Contents

Problem 11

Solution 1

Solution 2

Modulus Solution

See also