Difference between revisions of "2012 AMC 12A Problems/Problem 17"
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== Solution == | == Solution == | ||
− | 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> | + | 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> 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 == |
Revision as of 18:53, 7 February 2016
Problem
Let be a subset of
with the property that no pair of distinct elements in
has a sum divisible by
. What is the largest possible size of
?
Solution
Of the integers from to
, there are six each of
. We can create several rules to follow for the elements in subset
. No element can be
iff there is an element that is
. No element can be
if there is an element that is
. Thus we can pick 6 elements from either
or
and 6 elements from either
or
for a total of
elements. Considering
, there can be one element that is so because it will only be divisible by
if paired with another element that is
. The final answer is
.
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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