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

(Created page with "== Problem == Let <math>M</math> be the least common multiple of all the integers <math>10</math> through <math>30,</math> inclusive. Let <math>N</math> be the least common mu...")
 
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==Solution==
 
==Solution==
  
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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 <math>\text{lcm}</math> of . In this case, <math>M = 2^4 \cdot 3^3 \cdot 5^2 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29.</math> Now, with the same logic, we find that <math>N = M \cdot 2 \cdot 37,</math> because we have an extra power of <math>2</math> and an extra power of <math>37.</math> Thus, <math>\frac{N}{M} = 2\cdot 37 = 74</math>. Thus, our answer is <math>\boxed {\textbf{(E)}}.</math>
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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 <math>\text{lcm}</math> of. In this case, <math>M = 2^4 \cdot 3^3 \cdot 5^2 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29.</math> Now, with the same logic, we find that <math>N = M \cdot 2 \cdot 37,</math> because we have an extra power of <math>2</math> and an extra power of <math>37.</math> Thus, <math>\frac{N}{M} = 2\cdot 37 = 74</math>. Thus, our answer is <math>\boxed {\textbf{(E)}}.</math>
  
 
~NH14
 
~NH14

Revision as of 19:01, 23 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

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, with 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 $\boxed {\textbf{(E)}}.$

~NH14

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