2020 AMC 12A Problems/Problem 21

Problem 21

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

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 osas

2020 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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2020 AMC 12A Problems/Problem 21

Problem 21

Solution