Difference between revisions of "2022 AMC 10B Problems/Problem 8"
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~BrandonZhang202415 | ~BrandonZhang202415 | ||
+ | |||
+ | ==Solution 3== | ||
+ | We find a pattern. | ||
+ | <cmath>\begin{align*} | ||
+ | &\{1,2,3,\ldots,10\}, \\ | ||
+ | &\{11,12,13,\ldots,20\},\\ | ||
+ | &\{21,22,23,\ldots,30\},\\ | ||
+ | &\vdots\\ | ||
+ | &\{991,992,993,\ldots,1000\}. | ||
+ | \end{align*}</cmath> | ||
+ | Through quick listing <math>(7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98)</math>, 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: <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> | ||
+ | |||
== See Also == | == See Also == |
Revision as of 08:20, 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
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
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
![]()
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.