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

Revision as of 17:45, 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 $16!$ and 14 appears in $7!$ to $16!$ . So, there is an odd amount of 7's since there are 10 7's from $7!$ to $16!$ and 3 7's from $7!$ to $16!$ , and 10+3=13 which is odd. So we need to multiply by 7 to get a perfect square. Since 30 is not a divisor of 7, our answer is 70 which is $\boxed{\text{C}}$ .

~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)

@@ Line 32: / Line 32: @@
 We can immediately eliminate B, D, and E since 13 only appears in <math>13!, 14!, 15, 16!</math>, so <math>13\cdot 13\cdot 13\cdot 13</math> is a perfect square.
 Next, we can test if 7 is possible (and if it is not we can use process of elimination)
-appears in <math>7!</math> to <math>16!</math>
+appears in <math>7!</math> to <math>16!</math> and
-appears in <math>14!</math> to <math>16!</math>
+appears in <math>7!</math> to <math>16!</math>.
-So, there is an odd amount of 7's. Since 30 is not a divisor of 7, our answer is 70 which is <math>\boxed{\text{C}}</math>.
+So, there is an odd amount of 7's since there are 10 7's from <math>7!</math> to <math>16!</math> and 3 7's from <math>7!</math> to <math>16!</math>, and 10+3=13 which is odd. So we need to multiply by 7 to get a perfect square. Since 30 is not a divisor of 7, our answer is 70 which is <math>\boxed{\text{C}}</math>.
 ~aleyang

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