Difference between revisions of "2024 AMC 10A Problems/Problem 20"
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\textbf{(D) }654 \qquad | \textbf{(D) }654 \qquad | ||
\textbf{(E) }675 \qquad</math> | \textbf{(E) }675 \qquad</math> | ||
+ | =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> | ||
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==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 16:25, 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
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 |
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