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 set goes: <math>1,1,2,1,2,1,2,1,1,2.</math> We can reasonably conclude that the pattern <math>1,1,2,1,2,1,2</math> repeats every 7 times. So, for every 7 sets, there are three multiples of 7. We calculate the floor of 100/7 and multiply that by 3 (we disregard the remainder of two since it doesn't add any extra sets with 2 multiples of 7). We get <math>14*3= \boxed{\textbf{(B) }42}</math> | We see that the pattern for the number of multiples per set goes: <math>1,1,2,1,2,1,2,1,1,2.</math> We can reasonably conclude that the pattern <math>1,1,2,1,2,1,2</math> repeats every 7 times. So, for every 7 sets, there are three multiples of 7. We calculate the floor of 100/7 and multiply that by 3 (we disregard the remainder of two since it doesn't add any extra sets with 2 multiples of 7). We get <math>14*3= \boxed{\textbf{(B) }42}</math> | ||
| + | ==Video Solution 1== | ||
| + | https://youtu.be/PdyKJ1p9Y2w | ||
| + | |||
| + | ~Education, the Study of Everything | ||
== See Also == | == See Also == | ||
Revision as of 21:19, 19 November 2022
- The following problem is from both the 2022 AMC 10B #8 and 2022 AMC 12B #6, so both problems redirect to this page.
Problem
Consider the following 
sets of 
elements each:
How many of these sets contain exactly two multiples of 
?
Solution 1
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
Each set contains exactly 
or 
multiples of .
There are 
total sets and 
multiples of .
Thus, there are 
sets with multiples of .
~BrandonZhang202415
Solution 3
We find a pattern.
Through quick listing 
, we can figure out that the first set has 1 multiple of 7. The second set has 1 multiple of 7. The third set has 2 multiples of 7. The fourth set has 1 multiple of 7. The fifth set has 2 multiples of 7. The sixth set has 1 multiple of 7. The seventh set has 2 multiples of 7. The eighth set has 1 multiple of 7. The ninth set has 1 multiples of 7. The tenth set has 2 multiples of 7.
We see that the pattern for the number of multiples per set goes: 
We can reasonably conclude that the pattern 
repeats every 7 times. So, for every 7 sets, there are three multiples of 7. We calculate the floor of 100/7 and multiply that by 3 (we disregard the remainder of two since it doesn't add any extra sets with 2 multiples of 7). We get 
Video Solution 1
~Education, the Study of Everything
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. 
