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

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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, <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> Now, using the same logic, we find that <cmath>N = M \cdot 2 \cdot 37,</cmath> 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 = \boxed{\textbf{(D)}\ 74}.</math>
  
 
~NH14
 
~NH14
 +
==Video Solution by TheBeautyofMath==
 +
https://youtu.be/wlDlByKI7A8?t=410
 +
 +
~IceMatrix
 +
==See Also==
 +
 +
{{AMC12 box|year=2021 Fall|ab=A|num-b=7|num-a=9}}

Latest revision as of 21:57, 7 April 2022

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, 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, $\frac{N}{M} = 2\cdot 37 = \boxed{\textbf{(D)}\ 74}.$

~NH14

Video Solution by TheBeautyofMath

https://youtu.be/wlDlByKI7A8?t=410

~IceMatrix

See Also

2021 Fall AMC 12A (Problems • Answer Key • Resources)
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
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