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)== |
Latest revision as of 00:19, 19 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
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) 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.