2013 AIME I Problems/Problem 11
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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 such that when , , or 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 = ; this LCM is hereby denoted and .
1, 2, 3, 4, 5, 6, 7, 8, 10, and 12 all divide , so
We have the following three modulo equations:
To solve the equations, you can notice the answer must be of the form where is an integer.
This must be divisible by LCM, which is .
Therefore, , which is an integer. Factor out 3 and divide to get . Therefore, . We can use Bezout's Identity or a Euclidean Algorithm bash to solve for the least of and .
We find that the least is and the least is .
Since we want to factor , don't multiply; we already know that the prime factors of are , , , and , and since is prime, we have .
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 . This number is also divisible by 1, 2, 3, 4, 5, 6, 7, 8, 10, and 12, thus, the three numbers are .
Thus, when taken mod 9, 11, 13. Since is congruent to 6 mod 9 and 3 mod 13, and congruent to 8 mod 11, the number must be a number that is congruent to mod , mod (because is a multiple of , which is a factor of that can be divided out) and cause to become when multiplied under modulo 11.
Looking at the last condition shows that mod 11 (after a bit of bashing) and is congruent to mod and mod 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 elements where is the modulo being taken, thus, there must be a solution in these lists as adding to this solution yields the next smallest solution. In this case, is the solution for and thus the answer is . Since is prime, the sum of the prime factors is .
See also
2013 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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