Difference between revisions of "2022 AMC 10B Problems/Problem 8"
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<math>\textbf{(A)}\ 40\qquad\textbf{(B)}\ 42\qquad\textbf{(C)}\ 43\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 50</math> | <math>\textbf{(A)}\ 40\qquad\textbf{(B)}\ 42\qquad\textbf{(C)}\ 43\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 50</math> | ||
− | ==Solution 1 | + | ==Solution 1 (Casework)== |
We apply casework to this problem. The only sets that contain two multiples of seven are those for which: | We apply casework to this problem. The only sets that contain two multiples of seven are those for which: | ||
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~MRENTHUSIASM | ~MRENTHUSIASM | ||
− | ==Solution 2 | + | ==Solution 2 (Find A Pattern)== |
We find a pattern. | We find a pattern. | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
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~~Hayabusa1 | ~~Hayabusa1 | ||
+ | |||
==Video Solution by Interstigation== | ==Video Solution by Interstigation== | ||
https://youtu.be/_KNR0JV5rdI?t=884 | https://youtu.be/_KNR0JV5rdI?t=884 |
Revision as of 16:40, 11 November 2023
- The following problem is from both the 2022 AMC 10B #8 and 2022 AMC 12B #6, so both problems redirect to this page.
Contents
Problem
Consider the following sets of elements each: How many of these sets contain exactly two multiples of ?
Solution 1 (Casework)
We apply casework to this problem. The only sets that contain two multiples of seven are those for which:
- The multiples of are and That is, the first and eighth elements of such sets are multiples of
- The multiples of are and That is, the second and ninth elements of such sets are multiples of
- The multiples of are and That is, the third and tenth elements of such sets are multiples of
The first element is for some integer It is a multiple of when
The second element is for some integer It is a multiple of when
The third element is for some integer It is a multiple of when
Each case has sets. Therefore, the answer is
~MRENTHUSIASM
Solution 2 (Find A Pattern)
We find a pattern. We can figure out that the first set has multiple of . The second set also has multiple of . The third set has multiples of . The fourth set has multiple of . The fifth set has multiples of . The sixth set has multiple of . The seventh set has multiples of . Calculating this pattern further, we can see (reasonably) that it repeats for each sets. We see that the pattern for the number of multiples per sets goes: So, for every sets, there are three sets with multiples of . We calculate and multiply that by . (We also disregard the remainder of since it doesn't add any extra sets with multiples of .). We get .
~(edited by) ProProtractor
Solution 3 (Fastest)
Each set contains exactly or multiples of .
There are total sets and multiples of .
Thus, there are sets with multiples of .
~BrandonZhang202415
Video Solution (🚀Under 3 min🚀)
~Education, the Study of Everything
Video Solution(1-16)
~~Hayabusa1
Video Solution by Interstigation
https://youtu.be/_KNR0JV5rdI?t=884
See Also
2022 AMC 10B (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 10 Problems and Solutions |
2022 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 5 |
Followed by Problem 7 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.