Difference between revisions of "2020 AMC 12A Problems/Problem 21"

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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

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

2020 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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All AMC 12 Problems and Solutions

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

@@ Line 36: / Line 36: @@
 by Linty Huang
+== Solution 3 ==
+As in the previous solutions, we start with
+<cmath>\text{lcm}(5!,n) = 5\text{gcd}(10!,n)</cmath>
+From this we have that <math>\text{lcm}(5!,n) \,|\, 5\text{gcd}(10!,n)</math> , and in particular, <math>n \,|\, 5\text{gcd}(10!,n)</math>. However, <math>\text{gcd}(10!,n)\, |\, n</math>, so we must have <math>\text{gcd}(10!,n) = n</math> or <math>\text{gcd}(10!,n) = n/5</math>. If <math>\text{gcd}(10!,n) = n</math>, then we have <math>\text{lcm}(5!,n) = 5n</math>; because <math>5\, |\, 5!</math>, this implies that 5 does not divide <math>n</math>, so we must have <math>\text{gcd}(10!,n) = n/5</math>.
+Now we have <math>\text{lcm}(5!,n) = n</math>, implying that <math>5!\, |\, n</math>, and <math>n/5\, |\, 10!</math>. Writing out prime factorizations, this gives us
+<cmath>2^3 \cdot 3 \cdot 5 \,|\, n</cmath>
+<cmath>n \,|\, 2^8 \cdot 3^4 \cdot 5^2 \cdot 7</cmath>
+So <math>n</math> 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 <math>\text{gcd}(2^8 \cdot 3^4 \cdot 5^2 \cdot 7,n) = n/5</math> implies that <math>n</math> has 2 factors of 5. Thus, there are <math>6 \cdot 4 \cdot 2 = 48</math> possible choices for <math>n</math>, and our answer is <math>\boxed{\textbf{(D) 48}}</math>.
+-gumbymoo
 {{AMC12 box|year=2020|ab=A|num-b=20|num-a=22}}
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