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> 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>. | 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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| + | == Video Solution by OmegaLearn == | ||
| + | https://youtu.be/orrw4VydBTk?t=483 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
== See Also == | == See Also == | ||
Revision as of 07:37, 4 November 2022
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 
if 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 
.
Video Solution by OmegaLearn
https://youtu.be/orrw4VydBTk?t=483
~ pi_is_3.14
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
