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

(Solution)
(Solution)
Line 11: Line 11:
 
<cmath>
 
<cmath>
 
\begin{align*}
 
\begin{align*}
m \cdot 2 \cdot 4 \cdot 6 \cdot 8 \cdot 10 \cdot 12 \cdot 14 \cdot 16 &=
+
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 5  \cdot 7
 
\end{align*}
 
\end{align*}
 +
m &= 2 \cdot 5 \cdot 7 = 70
 
</cmath>
 
</cmath>

Revision as of 16:34, 15 November 2023

Problem

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

Solution

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 5  \cdot (2 \cdot 3)  \cdot 7  \cdot (2  \cdot 2 \cdot 2)\\
&\equiv m  \cdot 2 \cdot 5  \cdot 7
\end{align*}
m &= 2 \cdot 5 \cdot 7 = 70 (Error compiling LaTeX. Unknown error_msg)