Difference between revisions of "2022 AMC 10B Problems/Problem 8"
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<li>The multiples of <math>7</math> are <math>3\pmod{10}</math> and <math>0\pmod{10}.</math></li><p> | <li>The multiples of <math>7</math> are <math>3\pmod{10}</math> and <math>0\pmod{10}.</math></li><p> | ||
</ol> | </ol> | ||
| − | Each case has <math>\lfloor\frac{100}{7}\rfloor=14</math> sets. Therefore, the answer is <math>14\cdot3=\boxed{\textbf{(B)}\ 42}.</math> | + | Each case has <math>\left\lfloor\frac{100}{7}\right\rfloor=14</math> sets. Therefore, the answer is <math>14\cdot3=\boxed{\textbf{(B)}\ 42}.</math> |
~MRENTHUSIASM | ~MRENTHUSIASM | ||
Revision as of 16:08, 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.
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 
- The multiples of are
and 
- The multiples of are
and 
and 
and 
and 
Each case has 
sets. Therefore, the answer is 
~MRENTHUSIASM
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
