2016 AIME I Problems/Problem 12
Contents
Problem
Find the least positive integer such that is a product of at least four not necessarily distinct primes.
Solution 1
is the product of two consecutive integers, so it is always even. Thus is odd and never divisible by . Thus any prime that divides must divide . We see that . We can verify that is not a perfect square mod for each of . Therefore, all prime factors of are .
Let for primes . From here, we could go a few different ways:
Solution 1a
Suppose ; then . Reducing modulo 11, we get so .
Suppose . Then we must have , which leads to , i.e., .
leads to (impossible)! Then leads to , a prime (impossible). Finally, for we get .
Thus our answer is .
Solution 1b
Let for primes . If , then . We can multiply this by and complete the square to find . But hence we have pinned a perfect square strictly between two consecutive perfect squares, a contradiction. Hence . Thus , or . From the inequality, we see that . , so and we are done.
Solution 2
First, we can show that . This can be done by just testing all residue classes.
For example, we can test or to show that is not divisible by 2.
Case 1: m = 2k
Case 2: m = 2k+1
Now, we can test , which fails, so we test , and we get m = .
-AlexLikeMath
Video Solution
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See Also
2016 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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