Difference between revisions of "2022 AMC 10B Problems/Problem 8"
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&\{991,992,993,\ldots,1000\}. | &\{991,992,993,\ldots,1000\}. | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | |||
How many of these sets contain exactly two multiples of <math>7</math>? | How many of these sets contain exactly two multiples of <math>7</math>? | ||
<math>\textbf{(A)}\ 40\qquad\textbf{(B)}\ 42\qquad\textbf{(C)}\ 43\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 50</math> | <math>\textbf{(A)}\ 40\qquad\textbf{(B)}\ 42\qquad\textbf{(C)}\ 43\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 50</math> | ||
− | ==Solution== | + | ==Solution 1 (Casework)== |
− | 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;"> | ||
− | <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> That is, the first and eighth 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></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>3\pmod{10}</math> and <math>0\pmod{10}.</math></li><p> | + | <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> |
+ | 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 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> | ||
~MRENTHUSIASM | ~MRENTHUSIASM | ||
+ | |||
+ | ==Solution 2 (Find A Pattern)== | ||
+ | We find a pattern. | ||
+ | <cmath>\begin{align*} | ||
+ | &\{1,2,3,\ldots,10\}, \\ | ||
+ | &\{11,12,13,\ldots,20\},\\ | ||
+ | &\{21,22,23,\ldots,30\},\\ | ||
+ | &\vdots\\ | ||
+ | &\{991,992,993,\ldots,1000\}. | ||
+ | \end{align*}</cmath> | ||
+ | We can figure out that the first set has <math>1</math> multiple of <math>7</math>. The second set also has <math>1</math> multiple of <math>7</math>. The third set has <math>2</math> multiples of <math>7</math>. The fourth set has <math>1</math> multiple of <math>7</math>. The fifth set has <math>2</math> multiples of <math>7</math>. The sixth set has <math>1</math> multiple of <math>7</math>. The seventh set has <math>2</math> multiples of <math>7</math>. Calculating this pattern further, we can see (reasonably) that it repeats for each <math>7</math> sets. | ||
+ | We see that the pattern for the number of multiples per <math>7</math> sets goes: <math>1,1,2,1,2,1,2.</math> So, for every <math>7</math> sets, there are three sets with <math>2</math> multiples of <math>7</math>. We calculate <math>\left\lfloor\frac{100}{7}\right\rfloor</math> and multiply that by <math>3</math>. (We also disregard the remainder of <math>2</math> since it doesn't add any extra sets with <math>2</math> multiples of <math>7</math>.). We get <math>14\cdot3= \boxed{\textbf{(B) }42}</math>. | ||
+ | |||
+ | ~(edited by) mihikamishra | ||
+ | |||
+ | ==Solution 3 (Fastest)== | ||
+ | Each set contains exactly <math>1</math> or <math>2</math> multiples of <math>7</math>. | ||
+ | |||
+ | There are <math>\dfrac{1000}{10}=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>. | ||
+ | |||
+ | ~BrandonZhang202415 | ||
+ | |||
+ | ==Video Solution (🚀Under 3 min🚀)== | ||
+ | https://youtu.be/PdyKJ1p9Y2w | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | ==Video Solution(1-16)== | ||
+ | https://youtu.be/SCwQ9jUfr0g | ||
+ | |||
+ | ~~Hayabusa1 | ||
+ | |||
+ | ==Video Solution by Interstigation== | ||
+ | https://youtu.be/_KNR0JV5rdI?t=884 | ||
== See Also == | == See Also == |
Revision as of 00:19, 19 August 2024
- 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 1 (Casework)
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
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 third element is for some integer It is a multiple of when
Each case has sets. Therefore, the answer is
~MRENTHUSIASM
Solution 2 (Find A Pattern)
We find a pattern. We can figure out that the first set has multiple of . The second set also has multiple of . The third set has multiples of . The fourth set has multiple of . The fifth set has multiples of . The sixth set has multiple of . The seventh set has multiples of . Calculating this pattern further, we can see (reasonably) that it repeats for each sets. We see that the pattern for the number of multiples per sets goes: So, for every sets, there are three sets with multiples of . We calculate and multiply that by . (We also disregard the remainder of since it doesn't add any extra sets with multiples of .). We get .
~(edited by) mihikamishra
Solution 3 (Fastest)
Each set contains exactly or multiples of .
There are total sets and multiples of .
Thus, there are sets with multiples of .
~BrandonZhang202415
Video Solution (🚀Under 3 min🚀)
~Education, the Study of Everything
Video Solution(1-16)
~~Hayabusa1
Video Solution by Interstigation
https://youtu.be/_KNR0JV5rdI?t=884
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.