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Difference between revisions of "2023 AMC 10B Problems/Problem 15"

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==Problem==
 
==Problem==
 
What is the least positive integer <math>m</math> such that <math>m\cdot2!\cdot3!\cdot4!\cdot5!...16!</math> is a perfect square?
 
What is the least positive integer <math>m</math> such that <math>m\cdot2!\cdot3!\cdot4!\cdot5!...16!</math> is a perfect square?
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==Solution==
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== Solution 1 ==
  
 
Consider 2,
 
Consider 2,

Revision as of 16:36, 15 November 2023

Problem

What is the least positive integer $m$ such that $m\cdot2!\cdot3!\cdot4!\cdot5!...16!$ is a perfect square?

Solution 1

Consider 2, there are odd number of 2's in $2!\cdot3!\cdot4!\cdot5!...16!$ (We're not counting 3 2's in 8, 2 3's in 9, etc).

There are even number of 3's in $2!\cdot3!\cdot4!\cdot5!...16!$ ...

So, original expression reduce to \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)\\ &\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\\ m &= 2 \cdot 5 \cdot 7 = 70 \end{align*}

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