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

(Solution 1)
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<math>\textbf{(A)}\ 40\qquad\textbf{(B)}\ 42\qquad\textbf{(C)}\ 43\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 50</math>
 
<math>\textbf{(A)}\ 40\qquad\textbf{(B)}\ 42\qquad\textbf{(C)}\ 43\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 50</math>
  
==Solution 1 ~ Casework==
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==Solution 1 (Casework)==
 
We apply casework to this problem. The only sets that contain two multiples of seven are those for which:
 
We apply casework to this problem. The only sets that contain two multiples of seven are those for which:
  
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~MRENTHUSIASM
 
~MRENTHUSIASM
  
==Solution 2 ~ Find A Pattern==
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==Solution 2 (Find A Pattern)==
 
We find a pattern.  
 
We find a pattern.  
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
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~~Hayabusa1
 
~~Hayabusa1
 +
 
==Video Solution by Interstigation==
 
==Video Solution by Interstigation==
 
https://youtu.be/_KNR0JV5rdI?t=884
 
https://youtu.be/_KNR0JV5rdI?t=884

Revision as of 16:40, 11 November 2023

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 (Casework)

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 (Find A Pattern)

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*} We can figure out that the first set has $1$ multiple of $7$. The second set also 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 $7$. Calculating this pattern further, we can see (reasonably) that it repeats for each $7$ sets. We see that the pattern for the number of multiples per $7$ sets goes: $1,1,2,1,2,1,2.$ So, for every $7$ sets, there are three sets with $2$ multiples of $7$. We calculate $\left\lfloor\frac{100}{7}\right\rfloor$ and multiply that by $3$. (We also disregard the remainder of $2$ since it doesn't add any extra sets with $2$ multiples of $7$.). We get $14\cdot3= \boxed{\textbf{(B) }42}$.

~(edited by) ProProtractor

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🚀)

https://youtu.be/PdyKJ1p9Y2w

~Education, the Study of Everything

Video Solution(1-16)

https://youtu.be/SCwQ9jUfr0g

~~Hayabusa1

Video Solution by Interstigation

https://youtu.be/_KNR0JV5rdI?t=884

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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