2013 AIME I Problems/Problem 11
Contents
Problem
Ms. Math's kindergarten class has registered students. The classroom has a very large number, , of play blocks which satisfies the conditions:
(a) If , , or 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 satisfying the above conditions.
Solution 1
must be some multiple of ; this is hereby denoted and .
, , , , , , , , , and 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 , which is .
Therefore, , which is an integer. Factor out 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 , , and . The value of this can be found to be . This number is also divisible by , , , , , , , , , and , thus, the three numbers are .
Thus, when taken mod , , . Since is congruent to mod and mod , and congruent to mod , 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 .
Looking at the last condition shows that mod (after a bit of bashing) and is congruent to mod and mod as previously noted. Listing out the numbers congruent to mod and mod yield the following lists:
mod : , , , , , , , , , , ...
mod : , , , , , , , , , , , , ...
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 .
Solution 3
It is obvious that and so the only mod number of students are . Therefore, . Try some approaches and you will see that this one is one of the few successful ones:
Start by setting the two equations together, then we get . Divide by . Note that since the RHS is , and since is , then , where is some nonnegative integer, because must be .
This reduces to . Now, take out the With the same procedure, , where is some nonnegative integer.
You also get , at which point . cannot be equal to . Therefore, , and we know the prime factors of are so the answer is .
Solution 4
We start by noticing that for some integer in order to satisfy the first condition.
Next, we satisfy the second condition. Since must leave a remainder when dividing , they are not divisors of . Thus, we can eliminate all s.t. which leaves . Thus, . Now, we seek to find the least which satisfies this set of congruences.
By Chinese Remainder Theorem on the first two congruences, we find that (we divide by three before proceeding in the first congruence to ensure the minimal solution). Finally, by CRT again on and we find that .
Thus, the minimal value of is possible at . The prime factorization of this minimum value is and so the answer is .
Solution 5
As the problem stated, the number of boxes is definitely a multiple of , so we assume total number of boxes is
Then, according to statement, we get . So we have , we just write it to be Which tells that must be all odd number. Moreover, we can see can't be a multiple of (as is a multiple of ) which means that We let
Now, we write It is true that , let , it has Then, must be odd, let , it indicates Now, must be even, tells Eventually, must be even, , , is the smallest. This time,
So the number of balls is , the desired value is
~bluesoul
Solution 6(CRT Bash)
From part (a), we know that . From part (b), we know that . We can expand on part (a) by saying that for some . Rather than taking the three modulos together, we take them individually. The inverse of 2 mod 9 is easily seen to be .
Now moving to the second modulo which we leave as follows, Now the last modulo, CRT on the first and the third one results in . Now doing the second one and the one we just made, . Thus, the smallest value that works for . Thus
~YBSuburbanTea
See also
2013 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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