Difference between revisions of "2005 AMC 10B Problems/Problem 25"
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| + | ==Solution 1== | ||
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 <math>\boxed{\mathrm{(C)}\ 62}</math>. | ||
| − | + | ==Solution 2== | |
| − | "Cut" 125 into half. The maximum integer value in the smaller half is 62. Thus the answer is <math>\boxed{\mathrm{(C)}\62}</math>. | + | "Cut" <math>125</math> into half. The maximum integer value in the smaller half is <math>62</math>. Thus the answer is <math>\boxed{\mathrm{(C)}\ 62}</math>. |
== See Also == | == See Also == | ||
{{AMC10 box|year=2005|ab=B|num-b=24|after=Last Problem}} | {{AMC10 box|year=2005|ab=B|num-b=24|after=Last Problem}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 12:35, 6 July 2016
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 ?
Solution
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 
.
Solution 2
"Cut" into half. The maximum integer value in the smaller half is . Thus the answer is .
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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