Difference between revisions of "2024 AMC 10A Problems/Problem 20"
(→Problem) |
(→Solution 1) |
||
| Line 12: | Line 12: | ||
\textbf{(E) }675 \qquad</math> | \textbf{(E) }675 \qquad</math> | ||
=Solution 1= | =Solution 1= | ||
| − | By listing out the smallest possible elements of subset <math>S,</math> we can find that subset <math>S</math> starts with <math>\{1, 4, 8, 11, 14, 18, 21, 24, 28, 31, \dots\}.</math> It is easily noticed that the elements of the subset "loop around" every 3 elements, specifically adding 10 each time. This means that there will be <math>2024/10</math> or <math>202R4</math> whole loops in the subset <math>S,</math> implying that there will be <math>202*3 = 606</math> elements in S. However, we have undercounted, as we did not count the remainder that resulted from <math>2024/10</math><math>.</math> With a remainder of <math>4,</math> we can fit <math>2</math> more elements into the subset <math>S,</math> namely <math>2021</math> and <math>2024,</math> resulting in a total of <math>606+2</math> or <math>\boxed{\textbf{(C) }608}</math> elements in subset <math>S.</math> | + | By listing out the smallest possible elements of subset <math>S,</math> we can find that subset <math>S</math> starts with <math>\{1, 4, 8, 11, 14, 18, 21, 24, 28, 31, \dots\}.</math> It is easily noticed that the elements of the subset "loop around" every 3 elements, specifically adding 10 each time. This means that there will be <math>2024/10</math> or <math>202R4</math> whole loops in the subset <math>S,</math> implying that there will be <math>202*3 = 606</math> elements in S. However, we have undercounted, as we did not count the remainder that resulted from <math>2024/10</math><math>.</math> With a remainder of <math>4,</math> we can fit <math>2</math> more elements into the subset <math>S,</math> namely <math>2021</math> and <math>2024,</math> resulting in a total of <math>606+2</math> or <math>\boxed{\textbf{(C) }608}</math> elements in subset <math>S.</math> -weihou0 |
==See also== | ==See also== | ||
{{AMC10 box|year=2024|ab=A|before=19|num-a=21}} | {{AMC10 box|year=2024|ab=A|before=19|num-a=21}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 16:27, 8 November 2024
Problem
Let 
be a subset of 
such that the following two conditions hold:
- If 
and 
are distinct elements of , then 
- If and are distinct odd elements of , then 
.
What is the maximum possible number of elements in ?
Solution 1
By listing out the smallest possible elements of subset 
we can find that subset starts with 
It is easily noticed that the elements of the subset "loop around" every 3 elements, specifically adding 10 each time. This means that there will be 
or 
whole loops in the subset implying that there will be 
elements in S. However, we have undercounted, as we did not count the remainder that resulted from 
With a remainder of 
we can fit 
more elements into the subset namely 
and 
resulting in a total of 
or 
elements in subset 
-weihou0
See also
| 2024 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by 19 |
Followed by Problem 21 | |
| 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 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
