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

(→‎Solution)
(→‎Solution 2)
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We can prime factorize the solutions:
 
We can prime factorize the solutions:
−
A = 2 * 3 * 5
+
A = 2 \cdot 3 \cdot 5
−
B = 2 * 3 * 5 * 7 * 11 * 13
+
B = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13
−
C = 2 * 5 * 7
+
C = 2 \cdot 5 \cdot 7
−
D = 2 * 5 * 11 * 13
+
D = 2 \cdot 5 \cdot 11 \cdot 13
−
E = 7 * 11 * 13
+
E = 7 \cdot 11 \cot 13
  
 
We can immediately eliminated B, D, and E since 13 only appears in 13!, 14!, 15, 16!, so 13^4 is a perfect square.
 
We can immediately eliminated B, D, and E since 13 only appears in 13!, 14!, 15, 16!, so 13^4 is a perfect square.

Revision as of 16:38, 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*}


Solution 2

We can prime factorize the solutions: A = 2 \cdot 3 \cdot 5 B = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 C = 2 \cdot 5 \cdot 7 D = 2 \cdot 5 \cdot 11 \cdot 13 E = 7 \cdot 11 \cot 13

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

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