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

Revision as of 20:39, 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 $2!\cdot3!\cdot4!\cdot5!...16!$ ...etc,

So, we can reduce our original expression to $\begin{align*} m \cdot 2 \cdot 4 \cdot 6 \cdot 8 \cdot 10 \cdot 12 \cdot 14 \cdot 16 &\equiv m \cdot 2^8 \cdot (1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8)\\ &\equiv m \cdot 2 \cdot 3 \cdot (2 \cdot 2) \cdot 5 \cdot (2 \cdot 3) \cdot 7 \cdot (2 \cdot 2 \cdot 2)\\ &\equiv m \cdot 2 \cdot 5 \cdot 7\\ m &= 2 \cdot 5 \cdot 7 = 70 \end{align*}$

~Technodoggo ~minor edits by lucaswujc

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 $3! = 2! \cdot 3$ , $5! = 4! \cdot 5, ... 15! = 14! \cdot 15$ . So, $2!\cdot3! ={2!}^2\cdot3 \equiv 3, 4!\cdot5!={4!}^2\cdot5 \equiv 5, ... 14!\cdot15!={14!}^2\cdot15\equiv15$ . 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) & ~Technodoggo (add more examples)

Solution 4 (Bashy method)

We know that a perfect square must be in the form $2^{2a_1}\cdot3^{2a_2}\cdot5^{2a_3}...p^{2a_n}$ where $a_1, a_2, a_3, ..., a_n$ are nonnegative integers, and $p$ is the largest and $nth$ prime factor of our square number.

Let's assume $r=m\cdot2!\cdot3!\cdot4!\cdot5!...16!$ . We need to prime factorize $r$ and see which prime factors are raised to an odd power. Then, we can multiply one factor each of prime number with an odd number of factors to $m$ . We can do this by finding the number of factors of $2$ , $3$ , $5$ , $7$ , $11$ , and $13$ .

Case 1: Factors of $2$

We first count factors of $2^1$ in each of the factorials. We know there is one factor of $2^1$ each in $2!$ and $3!$ , two in $4!$ and $5!$ , and so on until we have $8$ factors of $2^1$ in $16!$ . Adding them all up, we have $1+1+2+2+...7+7+8=64$ .

Now, we count factors of $2^2$ in each of the factorials. We know there is one factor of $2^2$ each in $4!$ , $5!$ , $6!$ , and $7!$ , two in $8!$ , $9!$ , $10$ , and $11!$ , and so on until we have $4$ factors of $2^1$ in $16!$ . Adding them all up, we have $1+1+1+1+2+2+2+2+3+3+3+3+4=28$ .

Now we count factors of $2^3$ in each of the factorials. Using a similar method as above, we have a sum of $1+1+1+1+1+1+1+1+2=10$ .

Now we count factors of $2^4$ in each of the factorials. Using a similar method as above, we have a factor of $2^4$ in $16$ , so there is $1$ factor of $2^4$ .