Difference between revisions of "2022 AMC 10B Problems/Problem 8"
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~MRENTHUSIASM | ~MRENTHUSIASM | ||
+ | ==Solution 2== | ||
+ | Each set has exactly <math>1</math> or <math>2</math> multiples of <math>7</math>. | ||
+ | |||
+ | There are <math>\dfrac{991-1}{10}+1=100</math> total sets and <math>\left\lfloor\dfrac{1000}{7}\right\rfloor = 142</math> multiples of <math>7</math>. | ||
+ | |||
+ | Thus, there are <math>142-100=\boxed{\textbf{(B) }42}</math> sets with <math>2</math> multiples of <math>7</math>. | ||
== See Also == | == See Also == |
Revision as of 19:25, 17 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.
Contents
Problem
Consider the following sets of elements each:
How many of these sets contain exactly two multiples of ?
Solution
We apply casework to this problem:
- 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 second element is for some integer It is a multiple of when
Each case has sets. Therefore, the answer is
~MRENTHUSIASM
Solution 2
Each set has exactly or multiples of .
There are total sets and multiples of .
Thus, there are sets with multiples of .
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 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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.