2022 AMC 10B Problems/Problem 8: Difference between revisions
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==Solution 1== | ==Solution 1== | ||
We apply casework to this problem: | We apply casework to this problem. The only sets that contain two multiples of seven are those for which: | ||
<ol style="margin-left: 1.5em;"> | <ol style="margin-left: 1.5em;"> | ||
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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> | 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> | <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 | The third 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> | ||
</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 20:02, 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 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
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 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
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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
