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== |
− | 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 = | + | 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 |
Revision as of 16:12, 28 November 2021
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
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, Now, using the same logic, we find that because we have an extra power of and an extra power of Thus,
~NH14
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 |