Difference between revisions of "2021 Fall AMC 12A Problems/Problem 8"
(→Solution) |
Mathcosine (talk | contribs) (→Solution 2) |
||
| (7 intermediate revisions by 5 users not shown) | |||
| Line 6: | Line 6: | ||
==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 | ||
| + | |||
| + | ==Solution 2== | ||
| + | |||
| + | 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> | ||
| + | |||
| + | |||
| + | ~MathCosine | ||
| + | |||
| + | ==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 09:20, 22 October 2024
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, ![\[M = 2^4 \cdot 3^3 \cdot 5^2 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29.\]](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, 
~NH14
Solution 2
We know that between 32 and 40, there is one prime number, namely 
. This was not included in the previous , so we know that the value of 
will include a new 37. Then, let’s consider each individual number. ![\[32 = 2^5\]](http://latex.artofproblemsolving.com/5/d/8/5d89fdaaa49138dc832edcedad4a87e2d1663698.png)
has a new power of 2 now included in , so we add that to our list. ![\[33=3 \cdot 11\]](http://latex.artofproblemsolving.com/4/b/1/4b1d955b0c4a70efdce699f9d703d96400448cc9.png)
, which was included in the previous . Similarly, we can go through the numbers 32 to 40 and find that the only new powers of numbers added are ![\[2 \cdot 37 = 74\]](http://latex.artofproblemsolving.com/d/1/b/d1b808da2603240214dce96ab29401ee78b1ebec.png)
, which gives
~MathCosine
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 | |
