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

(→‎Solution)
(→‎Solution 2)
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<cmath>\text{lcm}{(5!, n)} = 5\text{gcd}{(10!, n)}.</cmath>
 
<cmath>\text{lcm}{(5!, n)} = 5\text{gcd}{(10!, n)}.</cmath>
−
Assume <math>\text{lcm}{(5!, n)}=k\cdot5!</math>, with some integer <math>k</math>. It follows that <math>k\cdot 4!=\text{gcd}{(10!, n)}</math>. It means that <math>n</math> has a factor <math>4!</math>. Since <math>n</math> is a multiple of <math>5</math>, <math>n</math> has a factor <math>5!</math>. Thus, <math>\text{lcm}{(5!, n)}=n=k\cdot5!</math>. The equation can be changed as
+
Assume <math>\text{lcm}{(5!, n)}=k\cdot5!</math>, with some integer <math>k</math>. It follows that <math>k\cdot 4!=\text{gcd}{(10!, n)}</math>. It means that <math>n</math> has a divisor <math>4!</math>. Since <math>n</math> is a multiple of <math>5</math>, <math>n</math> has a divisor <math>5!</math>. Thus, <math>\text{lcm}{(5!, n)}=n=k\cdot5!</math>. The equation can be changed as
 
<cmath>k\cdot5!=5\text{gcd}{(10!, k\cdot5!)}</cmath>
 
<cmath>k\cdot5!=5\text{gcd}{(10!, k\cdot5!)}</cmath>
 
<cmath>k=5\text{gcd}{(6\cdot7\cdot8\cdot9\cdot10, k)}</cmath>
 
<cmath>k=5\text{gcd}{(6\cdot7\cdot8\cdot9\cdot10, k)}</cmath>
Line 33: Line 33:
 
Similarly, <math>m</math> is a multiple of <math>5</math>, with a form of <math>5\cdot s</math>. We have  
 
Similarly, <math>m</math> is a multiple of <math>5</math>, with a form of <math>5\cdot s</math>. We have  
 
<cmath>s=\text{gcd}{(6\cdot7\cdot8\cdot9\cdot2, 5\cdot s)}=\text{gcd}{(2^5\cdot3^3\cdot7, s)}</cmath>
 
<cmath>s=\text{gcd}{(6\cdot7\cdot8\cdot9\cdot2, 5\cdot s)}=\text{gcd}{(2^5\cdot3^3\cdot7, s)}</cmath>
−
The equation holds, if <math>s</math> is a factor of <math>2^5\cdot3^3\cdot7</math>, which has <math>(5+1)(3+1)(1+1)=48</math> factors. The answer is  <math>\boxed{(D)48}</math>.
+
The equation holds, if <math>s</math> is a divisor of <math>2^5\cdot3^3\cdot7</math>, which has <math>(5+1)(3+1)(1+1)=\boxed{(D)48}</math> divisors.
 +
 
 +
by Linty Huang
  
 
{{AMC12 box|year=2020|ab=A|num-b=20|num-a=22}}
 
{{AMC12 box|year=2020|ab=A|num-b=20|num-a=22}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 07:12, 14 July 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

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 $(5+1)(3+1)(1+1)=\boxed{(D)48}$ divisors.

by Linty Huang

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

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