Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution)
(→‎Solution 2)
 
(14 intermediate revisions by 9 users not shown)
Line 1: Line 1:
−
==Problem 21==
+
==Problem==
  
 
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> and <math>n</math> equals <math>5</math> times the greatest common divisor of <math>10!</math> and <math>n?</math>
 
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> and <math>n</math> equals <math>5</math> times the greatest common divisor of <math>10!</math> and <math>n?</math>
Line 5: Line 5:
 
<math>\textbf{(A) } 12 \qquad \textbf{(B) } 24 \qquad \textbf{(C) } 36 \qquad \textbf{(D) } 48 \qquad \textbf{(E) } 72</math>
 
<math>\textbf{(A) } 12 \qquad \textbf{(B) } 24 \qquad \textbf{(C) } 36 \qquad \textbf{(D) } 48 \qquad \textbf{(E) } 72</math>
  
−
== Solution ==
+
== Solution 1==
  
 
We set up the following equation as the problem states:
 
We set up the following equation as the problem states:
Line 21: Line 21:
 
Thus, we can multiply the total possibilities of <math>n</math>'s factorization to determine the number of integers <math>n</math> which satisfy the equation, giving us <math>6 \times 4 \times 1 \times 2 = \boxed{\textbf{(D) } 48}</math>. ~ciceronii
 
Thus, we can multiply the total possibilities of <math>n</math>'s factorization to determine the number of integers <math>n</math> which satisfy the equation, giving us <math>6 \times 4 \times 1 \times 2 = \boxed{\textbf{(D) } 48}</math>. ~ciceronii
  
 +
== Solution 2==
  
 +
Like the Solution 1, we start from the equation:
 +
 +
<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 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=5\text{gcd}{(6\cdot7\cdot8\cdot9\cdot10, k)}</cmath>
 +
We can see that <math>k</math> is also a multiple of <math>5</math>, with a form of <math>5\cdot m</math>. Substituting it in the above equation, we have
 +
<cmath>m=5\text{gcd}{(6\cdot7\cdot8\cdot9\cdot2, m)}</cmath>
 +
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>
 +
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{(\textbf{D})48}</math> divisors. ~Linty Huang.shen k.kai
 +
 +
== 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
 +
 +
==Video Solution by Richard Rusczyk==
 +
https://www.youtube.com/watch?v=8S85536jpYw&list=PLyhPcpM8aMvJvwA2kypmfdtlxH90ShZCc&index=1&t=25s
 +
- AMBRIGGS
 +
 +
== Video Solution by OmegaLearn ==
 +
https://youtu.be/CWZkTCNu42o?t=846
 +
 +
~ pi_is_3.14
 +
 +
==See Also==
 
{{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}}

Latest revision as of 00:31, 21 August 2024

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 start 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. ~Linty Huang.shen k.kai

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

Video Solution by Richard Rusczyk

https://www.youtube.com/watch?v=8S85536jpYw&list=PLyhPcpM8aMvJvwA2kypmfdtlxH90ShZCc&index=1&t=25s - AMBRIGGS

Video Solution by OmegaLearn

https://youtu.be/CWZkTCNu42o?t=846

~ pi_is_3.14

See Also

2020 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12A_Problems/Problem_21&oldid=225465"