Difference between revisions of "2023 AMC 10B Problems/Problem 15"
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Consider 2, | Consider 2, | ||
there are odd number of 2's in <math>2!\cdot3!\cdot4!\cdot5!...16!</math> (We're not counting 3 2's in 8, 2 3's in 9, etc). | there are odd number of 2's in <math>2!\cdot3!\cdot4!\cdot5!...16!</math> (We're not counting 3 2's in 8, 2 3's in 9, etc). | ||
| + | |||
There are even number of 3's in <math>2!\cdot3!\cdot4!\cdot5!...16!</math> | There are even number of 3's in <math>2!\cdot3!\cdot4!\cdot5!...16!</math> | ||
... | ... | ||
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\begin{align*} | \begin{align*} | ||
m \cdot 2 \cdot 4 \cdot 6 \cdot 8 \cdot 10 \cdot 12 \cdot 14 \cdot 16 &\equiv m \cdot 2^8 \cdot (1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8)\\ | m \cdot 2 \cdot 4 \cdot 6 \cdot 8 \cdot 10 \cdot 12 \cdot 14 \cdot 16 &\equiv m \cdot 2^8 \cdot (1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8)\\ | ||
| − | &\equiv m \cdot 2 \cdot 3 \cdot 5 \cdot (2 \cdot 3) \cdot 7 \cdot (2 \cdot 2 \cdot 2)\\ | + | &\equiv m \cdot 2 \cdot 3 \cdot (2 \cdot 2) \cdot 5 \cdot (2 \cdot 3) \cdot 7 \cdot (2 \cdot 2 \cdot 2)\\ |
&\equiv m \cdot 2 \cdot 5 \cdot 7\\ | &\equiv m \cdot 2 \cdot 5 \cdot 7\\ | ||
m &= 2 \cdot 5 \cdot 7 = 70 | m &= 2 \cdot 5 \cdot 7 = 70 | ||
\end{align*} | \end{align*} | ||
</cmath> | </cmath> | ||
Revision as of 16:36, 15 November 2023
Problem
What is the least positive integer 
such that 
is a perfect square?
Solution
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
