2012 AMC 12A Problems/Problem 17

Problem

Let $S$ be a subset of $\{1,2,3,\dots,30\}$ with the property that no pair of distinct elements in $S$ has a sum divisible by $5$ . What is the largest possible size of $S$ ?

$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 13\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 16\qquad\textbf{(E)}\ 18$

Solution

Of the integers from $1$ to $30$ , there are six each of $0,1,2,3,4\ (\text{mod}\ 5)$ . We can create several rules to follow for the elements in subset $S$ . No element can be $1\ (\text{mod}\ 5)$ if there is an element that is $4\ (\text{mod}\ 5)$ . No element can be $2\ (\text{mod}\ 5)$ if there is an element that is $3\ (\text{mod}\ 5)$ . Thus we can pick 6 elements from either $1\ (\text{mod}\ 5)$ or $4\ (\text{mod}\ 5)$ and 6 elements from either $2\ (\text{mod}\ 5)$ or $3\ (\text{mod}\ 5)$ for a total of $6+6=12$ elements. Considering $0\ (\text{mod}\ 5)$ , there can be one element that is so because it will only be divisible by $5$ if paired with another element that is $0\ (\text{mod}\ 5)$ . The final answer is $\boxed{\textbf{(B)}\ 13}$ .

Video Solution by OmegaLearn

https://youtu.be/orrw4VydBTk?t=483

~ pi_is_3.14

2012 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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All AMC 12 Problems and Solutions

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2012 AMC 12A Problems/Problem 17

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Problem

Solution

Video Solution by OmegaLearn

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