# 2020 AMC 12A Problems/Problem 21

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## Problem

How many positive integers $n$ are there such that $n$ is a multiple of $5$, and the least common multiple of $5!$ and $n$ equals $5$ times the greatest common divisor of $10!$ and $n?$

$\textbf{(A) } 12 \qquad \textbf{(B) } 24 \qquad \textbf{(C) } 36 \qquad \textbf{(D) } 48 \qquad \textbf{(E) } 72$

## Solution 1

We set up the following equation as the problem states:

$$\text{lcm}{(5!, n)} = 5\text{gcd}{(10!, n)}.$$

Breaking each number into its prime factorization, we see that the equation becomes

$$\text{lcm}{(2^3\cdot 3 \cdot 5, n)} = 5\text{gcd}{(2^8\cdot 3^4 \cdot 5^2 \cdot 7, n)}.$$

We can now determine the prime factorization of $n$. We know that its prime factors belong to the set $\{2, 3, 5, 7\}$, as no factor of $10!$ has $11$ in its prime factorization, nor anything greater. Next, we must find exactly how many different possibilities exist for each.

There can be anywhere between $3$ and $8$ $2$'s and $1$ to $4$ $3$'s. However, since $n$ is a multiple of $5$, and we multiply the $\text{gcd}$ by $5$, there can only be $3$ $5$'s in $n$'s prime factorization. Finally, there can either $0$ or $1$ $7$'s.

Thus, we can multiply the total possibilities of $n$'s factorization to determine the number of integers $n$ which satisfy the equation, giving us $6 \times 4 \times 1 \times 2 = \boxed{\textbf{(D) } 48}$. ~ciceronii

## Solution 2

Like the Solution 1, we starts from the equation:

$$\text{lcm}{(5!, n)} = 5\text{gcd}{(10!, n)}.$$ Assume $\text{lcm}{(5!, n)}=k\cdot5!$, with some integer $k$. It follows that $k\cdot 4!=\text{gcd}{(10!, n)}$. It means that $n$ has a divisor $4!$. Since $n$ is a multiple of $5$, $n$ has a divisor $5!$. Thus, $\text{lcm}{(5!, n)}=n=k\cdot5!$. The equation can be changed as $$k\cdot5!=5\text{gcd}{(10!, k\cdot5!)}$$ $$k=5\text{gcd}{(6\cdot7\cdot8\cdot9\cdot10, k)}$$ We can see that $k$ is also a multiple of $5$, with a form of $5\cdot m$. Substituting it in the above equation, we have $$m=5\text{gcd}{(6\cdot7\cdot8\cdot9\cdot2, m)}$$ Similarly, $m$ is a multiple of $5$, with a form of $5\cdot s$. We have $$s=\text{gcd}{(6\cdot7\cdot8\cdot9\cdot2, 5\cdot s)}=\text{gcd}{(2^5\cdot3^3\cdot7, s)}$$ The equation holds, if $s$ is a divisor of $2^5\cdot3^3\cdot7$, which has $(5+1)(3+1)(1+1)=\boxed{(\textbf{D})48}$ divisors.

by Linty Huang

## Solution 3

As in the previous solutions, we start with

$$\text{lcm}(5!,n) = 5\text{gcd}(10!,n)$$

From this we have that $\text{lcm}(5!,n) \,|\, 5\text{gcd}(10!,n)$ , and in particular, $n \,|\, 5\text{gcd}(10!,n)$. However, $\text{gcd}(10!,n)\, |\, n$, so we must have $\text{gcd}(10!,n) = n$ or $\text{gcd}(10!,n) = n/5$. If $\text{gcd}(10!,n) = n$, then we have $\text{lcm}(5!,n) = 5n$; because $5\, |\, 5!$, this implies that 5 does not divide $n$, so we must have $\text{gcd}(10!,n) = n/5$.

Now we have $\text{lcm}(5!,n) = n$, implying that $5!\, |\, n$, and $n/5\, |\, 10!$. Writing out prime factorizations, this gives us

$$2^3 \cdot 3 \cdot 5 \,|\, n$$ $$n \,|\, 2^8 \cdot 3^4 \cdot 5^2 \cdot 7$$

So $n$ can have 3, 4, 5, 6, 7, or 8 factors of 2; 1, 2, 3, or 4 factors of two; and 0 or 1 factors of 7. Note that $\text{gcd}(2^8 \cdot 3^4 \cdot 5^2 \cdot 7,n) = n/5$ implies that $n$ has 2 factors of 5. Thus, there are $6 \cdot 4 \cdot 2 = 48$ possible choices for $n$, and our answer is $\boxed{\textbf{(D) 48}}$.

-gumbymoo

~ pi_is_3.14

## See Also

 2020 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 20 Followed byProblem 22 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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