# 2022 AMC 10B Problems/Problem 8

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. The only sets that contain two multiples of seven are those for which:

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 third 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

We find a pattern. \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*} Through quick listing $7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98$, we can figure out that the first set has $1$ multiple of $7$. The second set has $1$ multiple of $7$. The third set has $2$ multiples of $7$. The fourth set has $1$ multiple of $7$. The fifth set has $2$ multiples of $7$. The sixth set has $1$ multiple of $7$. The seventh set has $2$ multiples of $7$. The eighth set has $1$ multiple of $7$. The ninth set has $1$ multiples of $7$. The tenth set has $2$ multiples of $7$. We see that the pattern for the number of multiples per set goes: $1,1,2,1,2,1,2,1,1,2.$ We can reasonably conclude that the pattern $1,1,2,1,2,1,2$ repeats every $7$ times. So, for every $7$ sets, there are three multiples of $7$. We calculate $\left\lfloor\frac{100}{7}\right\rfloor$ and multiply that by $3$ (We disregard the remainder of $2$ since it doesn't add any extra sets with $2$ multiples of $7$.). We get $14\cdot3= \boxed{\textbf{(B) }42}$.

## Solution 3 (Fastest)

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

There are $\dfrac{1000}{10}=100$ 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

## Video Solution (🚀Under 3 min🚀)

~Education, the Study of Everything

~~Hayabusa1

## Video Solution by Interstigation

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