2022 AMC 10B Problems/Problem 8

Revision as of 16:22, 17 November 2022 by MRENTHUSIASM (talk | contribs)
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 first and eighth elements of such sets are multiples of $7.$ The first element is $1+10k$ for some integer $0\leq k\leq99.$

    The first element is a multiple of $7$ when $k=2,9,16,\ldots,93.$

  3. The multiples of $7$ are $2\pmod{10}$ and $9\pmod{10}.$
  4. The second and ninth elements of such sets are multiples of $7.$ The second element is $2+10k$ for some integer $0\leq k\leq99.$

    The second element is a multiple of $7$ when $k=4,11,18,\ldots,95.$

  5. The multiples of $7$ are $3\pmod{10}$ and $0\pmod{10}.$
  6. The third and tenth elements of such sets are multiples of $7.$ The second element is $3+10k$ for some integer $0\leq k\leq99.$

    The third element is a multiple of $7$ when $k=6,13,20,\ldots,97.$

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

~MRENTHUSIASM

See Also

2022 AMC 10B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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