Difference between revisions of "2024 AMC 10A Problems/Problem 20"
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\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> | + | All lists are organized in ascending order: |
| + | |||
| + | 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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| + | To prove that this is the best we can do, consider adding each element one by one, for the first element, say n. If n is greater than 2, we can choose n - 2 which is always better. Therefore, n = 1 or n = 2. | ||
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| + | If n = 2 was optimal, then choose it, then the set of usable numbers in <math>S</math> becomes 5 through 2024. We can transform the usable set of <math>S</math> to <math>Q</math> where <math>Q</math> contains the numbers 1 through 2020. Because we assumed n = 2 was optimal, we can choose n = 2 for the set <math>Q</math> too. Because every element in <math>Q</math> is 4 below the elements of <math>S</math>, choosing 2 in <math>Q</math> would mean choosing 6 in set <math>S</math>. By induction we see that our list would be {2,6,10,14,18,.....2022} which only gives 506 elements which is sub-optimal. | ||
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| + | Therefore, we can conclude that n = 1 is optimal. n = 1 means greedy algorithm, choose the first valid one as given in the first paragraph. | ||
| + | |||
| + | $-weihou0 | ||
==See also== | ==See also== | ||
{{AMC10 box|year=2024|ab=A|num-b=19|num-a=21}} | {{AMC10 box|year=2024|ab=A|num-b=19|num-a=21}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 20:10, 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
All lists are organized in ascending order:
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 .
To prove that this is the best we can do, consider adding each element one by one, for the first element, say n. If n is greater than 2, we can choose n - 2 which is always better. Therefore, n = 1 or n = 2.
If n = 2 was optimal, then choose it, then the set of usable numbers in becomes 5 through 2024. We can transform the usable set of to 
where contains the numbers 1 through 2020. Because we assumed n = 2 was optimal, we can choose n = 2 for the set too. Because every element in is 4 below the elements of , choosing 2 in would mean choosing 6 in set . By induction we see that our list would be {2,6,10,14,18,.....2022} which only gives 506 elements which is sub-optimal.
Therefore, we can conclude that n = 1 is optimal. n = 1 means greedy algorithm, choose the first valid one as given in the first paragraph.
$-weihou0
See also
| 2024 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 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. 
