Difference between revisions of "2023 AMC 10B Problems/Problem 15"
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− | We can immediately eliminate B, D, and E since 13 only appears in 13!, 14!, 15, 16!, so | + | We can immediately eliminate B, D, and E since 13 only appears in 13!, 14!, 15, 16!, so $13\cdot 13\cdot 13\cdot 13is a perfect square. |
Next, we can test if 7 is possible (and if it is not we can use process of elimination) | Next, we can test if 7 is possible (and if it is not we can use process of elimination) | ||
7 appears in 7! to 16! | 7 appears in 7! to 16! |
Revision as of 16:41, 15 November 2023
Problem
What is the least positive integer such that is a perfect square?
Solution 1
Consider 2, there are odd number of 2's in (We're not counting 3 2's in 8, 2 3's in 9, etc).
There are even number of 3's in ...
So, original expression reduce to
Solution 2
We can prime factorize the solutions:
We can immediately eliminate B, D, and E since 13 only appears in 13!, 14!, 15, 16!, so $13\cdot 13\cdot 13\cdot 13is a perfect square. Next, we can test if 7 is possible (and if it is not we can use process of elimination) 7 appears in 7! to 16! 14 appears in 14! to 16! So, there is an odd amount of 7's. Since 30 is not a divisor of 7, our answer is 70.
~aleyang