Difference between revisions of "2021 Fall AMC 12A Problems/Problem 8"

(Solution 2: Removed repetitive solution. Prof. Chen agreed to this through PM ...)
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~NH14
 
~NH14
 
== Solution 2 ==
 
First, we compute <math>M</math>. We have
 
<cmath>
 
\[
 
M = 2^4 \cdot 3^3 \cdot 5^2 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29.
 
\]
 
</cmath>
 
 
Second, we compute <math>N</math>. We have
 
<cmath>
 
\[
 
N = M \cdot 2 \cdot 37 .
 
\]
 
</cmath>
 
 
Therefore, <math>\frac{N}{M} = 2 \cdot 37 = 74</math>.
 
 
Therefore, the answer is <math>\boxed{\textbf{(D) }74}</math>.
 
 
~Steven Chen (www.professorchenedu.com)
 
 
  
 
==See Also==
 
==See Also==
  
 
{{AMC12 box|year=2021 Fall|ab=A|num-b=7|num-a=9}}
 
{{AMC12 box|year=2021 Fall|ab=A|num-b=7|num-a=9}}

Revision as of 00:49, 26 November 2021

Problem

Let $M$ be the least common multiple of all the integers $10$ through $30,$ inclusive. Let $N$ be the least common multiple of $M,32,33,34,35,36,37,38,39,$ and $40.$ What is the value of $\frac{N}{M}?$

$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 37 \qquad\textbf{(D)}\ 74 \qquad\textbf{(E)}\ 2886$

Solution 1

By the definition of least common mutiple, we take the greatest powers of the prime numbers of the prime factorization of all the numbers, that we are taking the $\text{lcm}$ of. In this case, \[M = 2^4 \cdot 3^3 \cdot 5^2 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29.\] Now, using the same logic, we find that \[N = M \cdot 2 \cdot 37,\] because we have an extra power of $2$ and an extra power of $37.$ Thus, $\frac{N}{M} = 2\cdot 37 = 74$. Thus, our answer is $\boxed {\textbf{(D)}}.$

~NH14

See Also

2021 Fall AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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