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== | + | ==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 <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> | 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> | ||
~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 20:38, 25 November 2021
Contents
Problem
Let 
be the least common multiple of all the integers 
through 
inclusive. Let 
be the least common multiple of 
and 
What is the value of 
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 
of. In this case, ![\[M = 2^4 \cdot 3^3 \cdot 5^2 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29.\]](http://latex.artofproblemsolving.com/1/c/b/1cb91d6ceadd5ba2c66b0a1fdde2782e934ac64d.png)
Now, using the same logic, we find that ![\[N = M \cdot 2 \cdot 37,\]](http://latex.artofproblemsolving.com/b/b/e/bbedd025873806341208e8bb12c360d0d63d80c9.png)
because we have an extra power of 
and an extra power of 
Thus, 
. Thus, our answer is 
~NH14
Solution 2
First, we compute . We have
![\[ M = 2^4 \cdot 3^3 \cdot 5^2 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29. \]](http://latex.artofproblemsolving.com/6/4/c/64c0e38a4e43a051b6b76d2c72332434bd62b904.png)
Second, we compute . We have
![\[ N = M \cdot 2 \cdot 37 . \]](http://latex.artofproblemsolving.com/f/b/d/fbd6c3671a6759c5d808e7eb23616fbbb78a06d6.png)
Therefore, 
.
Therefore, the answer is 
.
~Steven Chen (www.professorchenedu.com)
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 | |
