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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>
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XXX
 
 
   <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>
−
XXX <p>
 
 
   <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>
−
XXX <p>
 
 
</ol>
 
</ol>
 +
Each case has <math>\lfloor\frac{100}{7}\rfloor=14</math> sets. Therefore, the answer is <math>14\cdot3=\boxed{\textbf{(B)}\ 42}.</math>
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM

Revision as of 16:07, 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 $100$ sets of $10$ elements each: \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*}

How many of these sets contain exactly two multiples of $7$?

$\textbf{(A)}\ 40\qquad\textbf{(B)}\ 42\qquad\textbf{(C)}\ 43\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 50$

Solution

We apply casework to this problem:

  1. The multiples of $7$ are $1\pmod{10}$ and $8\pmod{10}.$

  2. The multiples of $7$ are $2\pmod{10}$ and $9\pmod{10}.$

  3. The multiples of $7$ are $3\pmod{10}$ and $0\pmod{10}.$

Each case has $\lfloor\frac{100}{7}\rfloor=14$ sets. Therefore, the answer is $14\cdot3=\boxed{\textbf{(B)}\ 42}.$

~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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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