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Difference between revisions of "2012 AMC 12A Problems/Problem 17"

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== Solution ==
 
== Solution ==
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Of the integers from <math>1</math> to <math>30</math>, there are six each of <math>0,1,2,3,4\ (\text{mod}\ 5)</math>. We can create several rules to follow for the elements in subset <math>S</math>. No element can be <math>1\ (\text{mod}\ 5)</math> iff there is an element that is <math>4\ (\text{mod}\ 5)</math>. No element can be <math>2\ (\text{mod}\ 5)</math> iff there is an element that is <math>3\ (\text{mod}\ 5)</math>. Ignoring those that are <math>0\ (\text{mod}\ 5)</math>, we can get a subset <math>S</math> with <math>12</math> elements. Considering <math>0\ (\text{mod}\ 5)</math>, there can be one element that is so because it will only be divisible by <math>5</math> if paired with another element that is <math>0\ (\text{mod}\ 5)</math>. The final answer is <math>\boxed{\textbf{(B)}\ 13}</math>.
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Of the integers from <math>1</math> to <math>30</math>, there are six each of <math>0,1,2,3,4\ (\text{mod}\ 5)</math>. We can create several rules to follow for the elements in subset <math>S</math>. No element can be <math>1\ (\text{mod}\ 5)</math> if there is an element that is <math>4\ (\text{mod}\ 5)</math>. No element can be <math>2\ (\text{mod}\ 5)</math> if there is an element that is <math>3\ (\text{mod}\ 5)</math>. Thus we can pick 6 elements from either <math>1\ (\text{mod}\ 5)</math> or <math>4\ (\text{mod}\ 5)</math> and 6 elements from either <math>2\ (\text{mod}\ 5)</math> or <math>3\ (\text{mod}\ 5)</math> for a total of <math>6+6=12</math> elements. Considering <math>0\ (\text{mod}\ 5)</math>, there can be one element that is so because it will only be divisible by <math>5</math> if paired with another element that is <math>0\ (\text{mod}\ 5)</math>. The final answer is <math>\boxed{\textbf{(B)}\ 13}</math>.
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== See Also ==
 
== See Also ==
  
 
{{AMC12 box|year=2012|ab=A|num-b=16|num-a=18}}
 
{{AMC12 box|year=2012|ab=A|num-b=16|num-a=18}}
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{{MAA Notice}}

Revision as of 22:15, 8 July 2016

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 $\boxed{\textbf{(B)}\ 13}$.

See Also

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