Difference between revisions of "2023 AMC 10B Problems/Problem 15"
(→Solution 2) |
(→Solution 2) |
||
Line 23: | Line 23: | ||
We can prime factorize the solutions: <math> | We can prime factorize the solutions: <math> | ||
− | A = 2 \cdot 3 \cdot 5 | + | A = 2 \cdot 3 \cdot 5, |
− | B = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 | + | B = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13, |
− | C = 2 \cdot 5 \cdot 7 | + | C = 2 \cdot 5 \cdot 7, |
− | D = 2 \cdot 5 \cdot 11 \cdot 13 | + | D = 2 \cdot 5 \cdot 11 \cdot 13, |
− | E = 7 \cdot 11 \cot 13 | + | E = 7 \cdot 11 \cot 13, |
</math> | </math> | ||
Revision as of 16:39, 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 eliminated B, D, and E since 13 only appears in 13!, 14!, 15, 16!, so 13^4 is 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