2023 AMC 10B Problems/Problem 15

Problem

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

$\textbf{(A) }30\qquad\textbf{(B) }30030\qquad\textbf{(C) }70\qquad\textbf{(D) }1430\qquad\textbf{(E) }1001$

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 ...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

Perfect squares have all of the powers in their prime factorization even. To evaluate $2!\cdot3!\cdot4!\cdot5!...16!$ we get the following:

$(2^{15}) \times (3^{14}) \times ((2^2)^{13}) \times (5^{12}) \times ((2 \times 3)^{11}) \times (7^{10}) \times ((2^3)^{9}) \times ((3^2)^8) \times \\ ((5 \times 2)^7) \times (11^6) \times (((2^2) \times 3)^5) \times (13^4) \times ((7 \times 2)^3) \times ((3 \times 5)^2) \times ((2^4)^1)$.

Taking all powers $\mod 2$ we get:

$(2^1) \times ((2 \times 3)^1) \times (2^1) \times ((5 \times 2)^1) \times (3^1) \times ((7 \times 2)^1)$

Simplifying again, we finally get:

$(2^1) \times (5^1) \times (7^1)$

To make all the powers left even, we need to multiply by $(2 \times 5 \times 7)$ which is $\boxed{\text{(C) }   70}$.

~darrenn.cp

Solution 3

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 $14!$ 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 $14!$ to $16!$ since 7 appears in 14 once, 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 4

First, we note that $3! = 2! \cdot 3$, $5! = 4! \cdot 5, ... 15! = 14! \cdot 15$. 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 5 (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 $2^4$.

Adding all the factors of $2$, we have $103$. Since $103$ is odd, $m$ has one factor of $2$.

Case 2: Factors of $3$

We use a similar method as in case 1. We first count factors of $3^1$. We obtain the sum $1+1+1+2+2+2+...4+4+4+5+5=50$.

We count factors of $3^2$. We obtain the sum $1+1+1+1+1+1+1+1=8$.

Adding all the factors of $3$, we have $58$. Since $58$ is even, $m$ has $0$ factors of $3$.

Case 3: Factors of $5$

We count the factors of $5^1$: $1+1+1+1+1+2+2+2+2+2+3+3=21$. Since $21$ is odd, $m$ has one factor of $5$.

Case 4: Factors of $7$

We count the factors of $7^1$: $1+1+1+1+1+1+1+2+2+2=13$. Since $13$ is odd, $m$ has one factor of $7$.

Case 5: Factors of $11$

We count the factors of $11^1$: $1+1+1+1+1+1=6$. Since $6$ is even, $m$ has $0$ factors of $11$.

Case 6: Factors of $13$

We count the factors of $13^1$: $1+1+1+1=4$. Since $4$ is even, $m$ has $0$ factors of $13$.

Multiplying out all our factors for $m$, we obtain $2\cdot5\cdot7=\boxed{\text{ (C) }70}$.

~arjken

Solution 6 (Answer Choices)

We see that all the answer choices are divisible by $10$ except for $1001$, and we also notice that the answer choices have $3$, $7$, $11$, or $13$ as a prime factor.

Testing, we see that $3$, $11$, $13$ have an even power in the product, so we have that all the other answer choices will not work.

Therefore we just have $\boxed{\text{ (C) }70}$.

Solution 7 (Fastest Intuition)

Notice that you can add the factor $1!$ to the expression to make it $m\cdot1!\cdot2!\cdot3!\cdot4!\cdot5!\cdot6!\cdot7!\cdot8!\cdot9!\cdot10!\cdot11!\cdot12!\cdot13!\cdot14!\cdot15!\cdot16!$

Every consecutive pair $n!\cdot(n+1)!$ can be simplified to $n+1$ (divide out $n!^2$).

Resulting is the product $m\cdot2\cdot4\cdot6\cdot8\cdot10\cdot12\cdot14\cdot16 = m\cdot2^8\cdot8!$

This is easily simplified to $m\cdot2\cdot5\cdot7 = m\cdot70$ when dividing out perfect squares.

This means that $m$ must have at minimun a factor of $70$ which gives answer choice $\boxed{\text{ (C) }70}$.

~coolishu

Solution 8(Pretty Quick similar to solution 7)

$16!\left(15!\right)$ this can be written as $16\left(15!\right)\left(15!\right)$


Thus


$16\left(15!\right)^{2}$

Now lets try

$16!15!14!$

$16\left(15\right)\left(14!\right)15\left(14!\right)\left(14!\right)$


thus $16\left(15\right)^{2}\left(14\right)^{3}$


we can continue this pattern until one to get $m\cdot2!\cdot3!\cdot4!\cdot5!...16! = 16\left(15\right)^{2}\left(14\right)^{3}\left(13\right)^{4}\left(12\right)^{5}\left(11\right)^{6}....\left(1\right)^{15}$

we can remove all the even powers of expression above to get

$16\left(14\right)^{3}\left(12\right)^{5}\left(10\right)^{7}\left(8\right)^{9}\left(6\right)^{11}\left(4\right)^{13}\left(2\right)^{15}$


we can see that all off the numbers are even and multiples of 2,3,5,7

the biggest possible number we should get is 2 times 3 times 5 times 7 which is 210 which means out anwer should be A or C 30 or 70.

Then look all the stuff with 7 which would only consist of ${14}^3$

which needs an extra 7 to give it company, so put answer must be divisble by 7 and less than 210

which leaves 70 or C as our answer

$\boxed{\text{ (C) }70}$.



Note: you can also just remove all pairs of two to get $16\left(14\right)\left(12\right)\left(10\right)\left(8\right)\left(6\right)\left(4\right)\left(2\right)$

Video Solution by MegaMath

https://www.youtube.com/watch?v=le0KSx3Cy-g&t=28s

~megahertz13

Video Solution 2 by OmegaLearn

https://youtu.be/fB81j1vbwdM

Video Solution 3 by SpreadTheMathLove

https://www.youtube.com/watch?v=sqVY5-h4vfo

Video Solution 4 by paixiao

https://www.youtube.com/watch?v=EvA2Nlb7gi4&t=238s

Video Solution

https://youtu.be/eW9eBpalm7I

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

See also

2023 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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All AMC 10 Problems and Solutions

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