2018 AMC 8 Problems/Problem 23

How many positive three-digit integers have a remainder of 2 when divided by 6, a remainder of 5 when divided by 9, and a remainder of 7 when divided by 11?

$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$

Solution

We will use constructive counting to solve this. There are $2$ cases: Either all $3$ points are adjacent, or exactly $2$ points are adjacent. If all $3$ points are adjacent, then we obviously have $8$ choices. If we have exactly $2$ adjacent points, then we will have $8$ places to put the adjacent points and also $4$ places to put the remaining point, so we have $8*4$ choices. The total amount of choices is ${8 \choose 3} = 8*7$ . Thus our answer is $\frac{8+8*4}{8*7}= \frac{1+4}{7}=\boxed{\textbf{(D) } \frac 57}$

Solution 2 (Complementary)

We can decide $2$ adjacent points with $8$ choices. The remaining point will have $6$ choices. However, we have counted the case with $3$ adjacent points twice, so we need to subtract this case once. The case with the $3$ adjacent points has $8$ arrangements, so our answer is $\frac{8*6-8}{8*7}=\boxed{\textbf{(D) } \frac 57}$

2018 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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2018 AMC 8 Problems/Problem 23

Solution

Solution 2 (Complementary)

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