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Difference between revisions of "2022 AMC 10B Problems/Problem 8"
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\end{align*}</cmath>
 
\end{align*}</cmath>
 
Through quick listing <math>(7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98)</math>, 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.
 
Through quick listing <math>(7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98)</math>, 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: <math>1,1,2,1,2,1,2,1,1,2.</math> We can reasonably conclude that the pattern <math>1,1,2,1,2,1,2</math> repeats every 7 times. So, for every 7 sets, there are three multiples of 7. We calculate the floor of 100/7 and multiply that by 3 (we disregard the remainder of two since it doesn't add any extra sets with 2 multiples of 7). We get <math>14*3= \boxed{\textbf{(B) }42}</math>
−
We see that the pattern for the number of multiples per set goes: <math>1,1,2,1,2,1,2,1,1,2.</math> We can reasonably conclude that the pattern <math>1,1,2,1,2,1,2</math> repeats every 7 times. So, for every 7 sets, there are three multiples of 7. We calculate the floor of 100/7 and multiply that by 3 (we disregard the remainder of two since it doesn't add any extra sets with 2 multiples of 7). We get <math>14*3= \boxed{\textbf{(B) }42}</math>
 
  
  

Revision as of 07:20, 19 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. 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 $k=2,9,16,\ldots,93.$

  2. 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 $7.$

    3. The second element is $2+10k$ for some integer $0\leq k\leq99.$ It is a multiple of $7$ when $k=4,11,18,\ldots,95.$

  3. 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 $7.$

    4. 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

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

Solution 3

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 the floor of 100/7 and multiply that by 3 (we disregard the remainder of two since it doesn't add any extra sets with 2 multiples of 7). We get $14*3= \boxed{\textbf{(B) }42}$


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

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