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

Revision as of 12:32, 2 September 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

We apply casework to this problem. The only sets that contain two multiples of seven are those for which:

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 $7.$

The first element is $1+10k$ for some integer $0\leq k\leq99.$ It is a multiple of $7$ when $k=2,9,16,\ldots,93.$

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 $7.$

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

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 $7.$

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 1

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

@@ Line 61: / Line 61: @@
 ==Video Solution by Interstigation==
 https://youtu.be/_KNR0JV5rdI?t=884
-~~(chimpionboy)(slight error) Can someone fix the pdf, it shows answer choices  A and B as 42
 == See Also ==
 {{AMC10 box|year=2022|ab=B|num-b=7|num-a=9}}
 {{AMC12 box|year=2022|ab=B|num-b=5|num-a=7}}
 {{MAA Notice}}

2022 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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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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