Difference between revisions of "2022 AMC 10B Problems/Problem 8"
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<li>The multiples of <math>7</math> are <math>1\pmod{10}</math> and <math>8\pmod{10}.</math></li><p> | <li>The multiples of <math>7</math> are <math>1\pmod{10}</math> and <math>8\pmod{10}.</math></li><p> | ||
| + | The first and eighth elements of such sets are multiples of <math>7.</math> The first element is <math>1+10k</math> for some integer <math>0\leq k\leq99.</math> <p> | ||
| + | The first element is a multiple of <math>7</math> when <math>k=2,9,16,\ldots,93.</math> | ||
<li>The multiples of <math>7</math> are <math>2\pmod{10}</math> and <math>9\pmod{10}.</math></li><p> | <li>The multiples of <math>7</math> are <math>2\pmod{10}</math> and <math>9\pmod{10}.</math></li><p> | ||
| + | The second and ninth elements of such sets are multiples of <math>7.</math> The second element is <math>2+10k</math> for some integer <math>0\leq k\leq99.</math> <p> | ||
| + | The second element is a multiple of <math>7</math> when <math>k=4,11,18,\ldots,95.</math> | ||
<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> | ||
| + | The third and tenth elements of such sets are multiples of <math>7.</math> The second element is <math>3+10k</math> for some integer <math>0\leq k\leq99.</math> <p> | ||
| + | The third element is a multiple of <math>7</math> when <math>k=6,13,20,\ldots,97.</math> | ||
</ol> | </ol> | ||
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> | 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> | ||
Revision as of 16:22, 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 
The first and eighth elements of such sets are multiples of 
The first element is 
for some integer 
The first element is a multiple of when 
and 
The second and ninth elements of such sets are multiples of The second element is 
for some integer
The second element is a multiple of when 
and 
The third and tenth elements of such sets are multiples of The second element is 
for some integer
The third element is a multiple of when 
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 | ||
| 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. 
