Difference between revisions of "2022 AMC 10B Problems/Problem 8"
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We see that the pattern for the number of multiples per <math>7</math> sets goes: <math>1,1,2,1,2,1,2.</math> So, for every <math>7</math> sets, there are three sets with <math>2</math> multiples of <math>7</math>. We calculate <math>\left\lfloor\frac{100}{7}\right\rfloor</math> and multiply that by <math>3</math>. (We also disregard the remainder of <math>2</math> since it doesn't add any extra sets with <math>2</math> multiples of <math>7</math>.). We get <math>14\cdot3= \boxed{\textbf{(B) }42}</math>. | We see that the pattern for the number of multiples per <math>7</math> sets goes: <math>1,1,2,1,2,1,2.</math> So, for every <math>7</math> sets, there are three sets with <math>2</math> multiples of <math>7</math>. We calculate <math>\left\lfloor\frac{100}{7}\right\rfloor</math> and multiply that by <math>3</math>. (We also disregard the remainder of <math>2</math> since it doesn't add any extra sets with <math>2</math> multiples of <math>7</math>.). We get <math>14\cdot3= \boxed{\textbf{(B) }42}</math>. | ||
| − | ~(edited by) | + | ~(edited by) mihikamishra |
==Solution 3 (Fastest)== | ==Solution 3 (Fastest)== | ||
Revision as of 23:19, 18 August 2024
- 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
and 
That is, the first and eighth elements of such sets are multiples of 
The first element is 
for some integer 
It is a multiple of when 
and 
That is, the second and ninth elements of such sets are multiples of
The second element is 
for some integer It is a multiple of when 
and 
That is, the third and tenth elements of such sets are multiples of
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) mihikamishra
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. 
