Difference between revisions of "2022 AMC 10B Problems/Problem 8"

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==Solution 2==
 
==Solution 2==
Each set has exactly <math>1</math> or <math>2</math> multiples of <math>7</math>.
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Each set contains exactly <math>1</math> or <math>2</math> multiples of <math>7</math>.
  
 
There are <math>\dfrac{991-1}{10}+1=100</math> total sets and <math>\left\lfloor\dfrac{1000}{7}\right\rfloor = 142</math> multiples of <math>7</math>.
 
There are <math>\dfrac{991-1}{10}+1=100</math> total sets and <math>\left\lfloor\dfrac{1000}{7}\right\rfloor = 142</math> multiples of <math>7</math>.

Revision as of 19:32, 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 1

We apply casework to this problem:

  1. The multiples of $7$ are $1\pmod{10}$ and $8\pmod{10}.$ That is, the first and eighth elements of such sets are multiples of $7.$
  2. The first element is $1+10k$ for some integer $0\leq k\leq99.$ It is a multiple of $7$ when $k=2,9,16,\ldots,93.$

  3. The multiples of $7$ are $2\pmod{10}$ and $9\pmod{10}.$ That is, the second and ninth elements of such sets are multiples of $7.$
  4. The second element is $2+10k$ for some integer $0\leq k\leq99.$ It is a multiple of $7$ when $k=4,11,18,\ldots,95.$

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

Solution 2

Each set contains exactly $1$ or $2$ multiples of $7$.

There are $\dfrac{991-1}{10}+1=100$ total sets and $\left\lfloor\dfrac{1000}{7}\right\rfloor = 142$ multiples of $7$.

Thus, there are $142-100=\boxed{\textbf{(B) }42}$ sets with $2$ multiples of $7$.

~BrandonZhang202415

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 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 12 Problems and Solutions

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