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

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<math>\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 37 \qquad\textbf{(D)}\ 74 \qquad\textbf{(E)}\ 2886</math>
 
<math>\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 37 \qquad\textbf{(D)}\ 74 \qquad\textbf{(E)}\ 2886</math>
  
==Solution 1==
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==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 <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 = 74</math>. Thus, our answer is <math>\boxed {\textbf{(D)}}.</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
  
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==Solution 2==
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We know that between 32 and 40, there is one prime number, namely <math>37</math>. This was not included in the previous <math>M</math>, so we know that the value of <math>\frac{N}{M}</math> will include a new 37. Then, let’s consider each individual number. <cmath>32 = 2^5</cmath> has a new power of 2 now included in <math>M</math>, so we add that to our list. <cmath>33=3 \cdot 11</cmath>, which was included in the previous <math>M</math>. Similarly, we can go through the numbers 32 to 40 and find that the only new powers of numbers added are <cmath>2 \cdot 37 = 74</cmath>, which gives <math>\frac{N}{M} = 2\cdot 37 = \boxed{\textbf{(D)}\ 74}.</math>
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~MathCosine
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==Video Solution by TheBeautyofMath==
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https://youtu.be/wlDlByKI7A8?t=410
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~IceMatrix
 
==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}}

Latest revision as of 09:20, 22 October 2024

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

Solution 2

We know that between 32 and 40, there is one prime number, namely $37$. This was not included in the previous $M$, so we know that the value of $\frac{N}{M}$ will include a new 37. Then, let’s consider each individual number. \[32 = 2^5\] has a new power of 2 now included in $M$, so we add that to our list. \[33=3 \cdot 11\], which was included in the previous $M$. Similarly, we can go through the numbers 32 to 40 and find that the only new powers of numbers added are \[2 \cdot 37 = 74\], which gives $\frac{N}{M} = 2\cdot 37 = \boxed{\textbf{(D)}\ 74}.$


~MathCosine

Video Solution by TheBeautyofMath

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

~IceMatrix

See Also

2021 Fall AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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All AMC 12 Problems and Solutions