Difference between revisions of "2005 AMC 10B Problems/Problem 25"
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The question asks for the maximum possible number of elements. The integers from <math>1</math> to <math>24</math> can be included because you cannot make <math>125</math> with integers from <math>1</math> to <math>24</math> without the other number being greater than <math>100</math>. The integers from <math>25</math> to <math>100</math> are left. They can be paired so the sum is <math>125</math>: <math>25+100</math>, <math>26+99</math>, <math>27+98</math>, <math>\ldots</math>, <math>62+63</math>. That is <math>38</math> pairs, and at most one number from each pair can be included in the set. The total is <math>24 + 38 = \boxed{\mathrm{(C)}\ 62}</math>. | The question asks for the maximum possible number of elements. The integers from <math>1</math> to <math>24</math> can be included because you cannot make <math>125</math> with integers from <math>1</math> to <math>24</math> without the other number being greater than <math>100</math>. The integers from <math>25</math> to <math>100</math> are left. They can be paired so the sum is <math>125</math>: <math>25+100</math>, <math>26+99</math>, <math>27+98</math>, <math>\ldots</math>, <math>62+63</math>. That is <math>38</math> pairs, and at most one number from each pair can be included in the set. The total is <math>24 + 38 = \boxed{\mathrm{(C)}\ 62}</math>. | ||
− | Also, it is possible to see that since the numbers <math>1</math> to <math>24</math> are in the set there are only the numbers <math>25</math> to <math>100</math> to consider. As <math>62+63</math> gives <math>125</math>, the numbers <math>25</math> to <math>62</math> can be put in subset <math>B</math> without having two numbers add up to <math>125</math>. In this way, subset <math>B</math> will have the numbers <math>1</math> to <math>62</math>, and so <math>\boxed{\mathrm{(C)}\ 62}</math>. | + | Also, it is possible to see that since the numbers <math>1</math> to <math>24</math> are in the set there are only the numbers <math>25</math> to <math>100</math> to consider. As <math>62+63</math> gives <math>125</math>, the numbers <math>25</math> to <math>62</math> can be put in subset <math>B</math> without having two numbers add up to <math>125</math>. In this way, subset <math>B</math> will have the numbers <math>1</math> to <math>62</math>, and so the answer is <math>\boxed{\mathrm{(C)}\ 62}</math>. |
====Solution 1 Alternate Solution==== | ====Solution 1 Alternate Solution==== |
Latest revision as of 06:51, 1 June 2021
Contents
Problem
A subset of the set of integers from to , inclusive, has the property that no two elements of sum to . What is the maximum possible number of elements in ?
-Solutions-
Solution 1
The question asks for the maximum possible number of elements. The integers from to can be included because you cannot make with integers from to without the other number being greater than . The integers from to are left. They can be paired so the sum is : , , , , . That is pairs, and at most one number from each pair can be included in the set. The total is . Also, it is possible to see that since the numbers to are in the set there are only the numbers to to consider. As gives , the numbers to can be put in subset without having two numbers add up to . In this way, subset will have the numbers to , and so the answer is .
Solution 1 Alternate Solution
Since there are 38 numbers that sum to , there are numbers not summing to ~mathboy282
Solution 2 (If you have no time)
"Cut" into half. The maximum integer value in the smaller half is . Thus the answer is .
Solution 3
The maximum possible number of elements includes the smallest numbers. So, subset where n is the maximum number of elements in subset . So, we have to find two consecutive numbers, and , whose sum is . Setting up our equation, we have . When we solve for , we get . Thus, the anser is .
~GentleTiger
See Also
2005 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
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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