Difference between revisions of "2022 AMC 10B Problems/Problem 8"
MRENTHUSIASM (talk | contribs) |
MRENTHUSIASM (talk | contribs) |
||
| Line 19: | Line 19: | ||
<ol style="margin-left: 1.5em;"> | <ol style="margin-left: 1.5em;"> | ||
| − | <li>The multiples of <math>7</math> are <math>1\pmod{10}</math> and <math>8\pmod{10}.</math> | + | <li>The multiples of <math>7</math> are <math>1\pmod{10}</math> and <math>8\pmod{10}.</math> That is, the first and eighth elements of such sets are multiples of <math>7.</math></li><p> |
| − | + | The first element is <math>1+10k</math> for some integer <math>0\leq k\leq99.</math> It 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> That is, the second and ninth elements of such sets are multiples of <math>7.</math></li><p> | |
| − | <li>The multiples of <math>7</math> are <math>2\pmod{10}</math> and <math>9\pmod{10}.</math> | + | The second element is <math>2+10k</math> for some integer <math>0\leq k\leq99.</math> It 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> That is, the third and tenth elements of such sets are multiples of <math>7.</math></li><p> | |
| − | + | The second element is <math>3+10k</math> for some integer <math>0\leq k\leq99.</math> It is a multiple of <math>7</math> when <math>k=6,13,20,\ldots,97.</math> | |
| − | <li>The multiples of <math>7</math> are <math>3\pmod{10}</math> and <math>0\pmod{10}.</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 17:24, 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 
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
and 
That is, the first and eighth elements of such sets are multiples of 
The first element is 
for some integer 
It is a multiple of when 
and 
That is, the second and ninth elements of such sets are multiples of
The second element is 
for some integer It is a multiple of when 
and 
That is, the third and tenth elements of such sets are multiples of
The second element is 
for some integer It 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. 
