2016 AIME I Problems/Problem 12
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 greater than or equal to .
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
Let , then . We can see for to have a second factor of 11. Let , we get , so . -Mathdummy
Solution 3
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
See Also
2016 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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