Difference between revisions of "2022 AMC 10B Problems/Problem 8"
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~FIREDRAGONMATH16 | ~FIREDRAGONMATH16 | ||
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+ | ==Solution 6 (reasonable count)== | ||
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+ | Note that this system loops every cycle of length <math>70</math>, or <math>7</math> such sets. From <math>1</math> to <math>70</math>, the multiples of <math>7</math> are <math>7</math>, <math>14</math>, <math>21</math>, <math>28</math>, <math>35</math>, <math>42</math>, <math>49</math>, <math>56</math>, <math>63</math>, and <math>70</math>; note that <math>(21,28),(42,49),</math> and <math>(63,70)</math> are in the same sets. Thus, for every <math>7</math> sets, we have <math>3</math> sets with exactly two multiples of <math>7</math>. We have <math>100</math> sets, which is <math>98+2=7\cdot14+2</math>; the first <math>98</math> sets contain <math>\dfrac37(7\cdot14)=42</math> desired sets. The last two sets comprise the integers from <math>981</math> to <math>999</math>; the multiples of <math>7</math> here are <math>987</math> and <math>994</math>. Neither of the two last sets contain two multiples of <math>7</math>, so our answer is simply <math>\boxed{\textbf{(D) }42</math>. | ||
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+ | ~Technodoggo | ||
==Video Solution (🚀Under 3 min🚀)== | ==Video Solution (🚀Under 3 min🚀)== |
Revision as of 22:50, 1 November 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
[hide]- 1 Problem
- 2 Solution 1 (Casework)
- 3 Solution 2 (Find A Pattern)
- 4 Solution 3 (Fastest)
- 5 Solution 4 (Simple Counting, Similar to Solution 1)
- 6 Solution 5 (Very Fast System of Equations)
- 7 Solution 6 (reasonable count)
- 8 Video Solution (🚀Under 3 min🚀)
- 9 Video Solution(1-16)
- 10 Video Solution by Interstigation
- 11 See Also
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 multiple of . The eighth set has multiples of . Disregarding the first set and then 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 (again, disregarding the first set) goes: So, for every sets after the first, 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
~(edited by) MiniGlasses2009
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
Solution 4 (Simple Counting, Similar to Solution 1)
Consecutive multiples of must differ by . So, if a set contains two multiples of , they must end with the digits and , and , or and . This reduces the problem to counting the amount of multiples of less than that end with , , and .
The first multiple of that ends with is . The next multiple that ends with occurs later, since that is the smallest multiple of we can add to without affecting the last digit. The greatest number of 's we can add to while keeping it less than is , because . Therefore, the set of multiples of less than ending with is , meaning there are of these particular multiples. We can use the same reasoning to count the multiples of that end with and .
The first multiple of that ends with is . The greatest number of 's we can add to here is also , since . The set of multiples of less than ending with is , giving multiples.
The first multiple of that ends with is . The greatest number of 's we can add to here is yet again , since . The set of multiples of less than ending with is , giving another multiples.
In total, there are of these multiples, and so sets with two multiples of .
~marsus16112
Solution 5 (Very Fast System of Equations)
Let be the number of sets with multiple of and be the number of sets with multiples of . Note that it is impossible for a set to have more than two multiples of .
Since there are a total of sets, . Also, since there are multiples of between and , we must have .
Solving the system of equations
for gives us the answer of .
~FIREDRAGONMATH16
Solution 6 (reasonable count)
Note that this system loops every cycle of length , or such sets. From to , the multiples of are , , , , , , , , , and ; note that and are in the same sets. Thus, for every sets, we have sets with exactly two multiples of . We have sets, which is ; the first sets contain desired sets. The last two sets comprise the integers from to ; the multiples of here are and . Neither of the two last sets contain two multiples of , so our answer is simply $\boxed{\textbf{(D) }42$ (Error compiling LaTeX. Unknown error_msg).
~Technodoggo
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.