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

(Created page with "==Problem 21== How many positive integers <math>n</math> are there such that <math>n</math> is a multiple of <math>5</math>, and the least common multiple of <math>5!</math>...")
 
Line 9: Line 9:
 
We set up the following equation as the problem states:
 
We set up the following equation as the problem states:
  
−
<cmath> \text{lcm}{(5!, n)} = 5\text{gcd}{(10!, n)}</cmath>.
+
<cmath> \text{lcm}{(5!, n)} = 5\text{gcd}{(10!, n)}.</cmath>
  
 
Breaking each number into its prime factorization, we see that the equation becomes
 
Breaking each number into its prime factorization, we see that the equation becomes
  
−
<cmath> \text{lcm}{(2^3\cdot 3 \cdot 5, n)} = 5\text{gcd}{(2^8\cdot 3^4 \cdot 5^2 \cdot 7, n)}</cmath>.
+
<cmath> \text{lcm}{(2^3\cdot 3 \cdot 5, n)} = 5\text{gcd}{(2^8\cdot 3^4 \cdot 5^2 \cdot 7, n)}.</cmath>
  
 
We can now determine the prime factorization of <math>n</math>. We know that its prime factors belong to the set <math>\{2, 3, 5, 7\}</math>, as no factor of <math>10!</math> has <math>11</math> in its prime factorization, nor anything greater. Next, we must find exactly how many different possibilities exist for each.
 
We can now determine the prime factorization of <math>n</math>. We know that its prime factors belong to the set <math>\{2, 3, 5, 7\}</math>, as no factor of <math>10!</math> has <math>11</math> in its prime factorization, nor anything greater. Next, we must find exactly how many different possibilities exist for each.

Revision as of 17:46, 1 February 2020

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

2020 AMC 12A (Problems • Answer Key • Resources)
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