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

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~aleyang
 
~aleyang
  
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== Solution 2 ==
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== Solution 3 ==
  
 
First, we note <math>3! = 2! \cdot 3</math>. Simplifying the whole sequence and cancelling out the squares, we get <math>3 \cdot 5 \cdot 7 \cdot 9 \cdot 11 \cdot 13 \cdot 15 \cdot 16!</math>. Prime factoring <math>16!</math> and cancelling out the squares, the only numbers that remain are <math>2, 5,</math> and <math>7</math>. Since we need to make this a perfect square, <math>m = 2 \cdot 5 \cdot 7</math>. Multiplying this out, we get <math>\boxed{\text{(C )}  70}</math>.
 
First, we note <math>3! = 2! \cdot 3</math>. Simplifying the whole sequence and cancelling out the squares, we get <math>3 \cdot 5 \cdot 7 \cdot 9 \cdot 11 \cdot 13 \cdot 15 \cdot 16!</math>. Prime factoring <math>16!</math> and cancelling out the squares, the only numbers that remain are <math>2, 5,</math> and <math>7</math>. Since we need to make this a perfect square, <math>m = 2 \cdot 5 \cdot 7</math>. Multiplying this out, we get <math>\boxed{\text{(C )}  70}</math>.
  
 
~yourmomisalosinggame (a.k.a. Aaron)
 
~yourmomisalosinggame (a.k.a. Aaron)

Revision as of 16:43, 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: $A = 2 \cdot 3 \cdot 5, B = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13, C = 2 \cdot 5 \cdot 7, D = 2 \cdot 5 \cdot 11 \cdot 13, E = 7 \cdot 11 \cdot 13,$

We can immediately eliminate B, D, and E since 13 only appears in $13!, 14!, 15, 16!$, so $13\cdot 13\cdot 13\cdot 13$ 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 which is $\boxed{\text{C 70}}$.

~aleyang

Solution 3

First, we note $3! = 2! \cdot 3$. Simplifying the whole sequence and cancelling out the squares, we get $3 \cdot 5 \cdot 7 \cdot 9 \cdot 11 \cdot 13 \cdot 15 \cdot 16!$. Prime factoring $16!$ and cancelling out the squares, the only numbers that remain are $2, 5,$ and $7$. Since we need to make this a perfect square, $m = 2 \cdot 5 \cdot 7$. Multiplying this out, we get $\boxed{\text{(C )} 70}$.

~yourmomisalosinggame (a.k.a. Aaron)

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