Difference between revisions of "2021 Fall AMC 12A Problems/Problem 8"
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==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 | ||
+ | ==Video Solution by TheBeautyofMath== | ||
+ | https://youtu.be/wlDlByKI7A8?t=410 | ||
+ | ~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 21:57, 7 April 2022
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
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 |