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Difference between revisions of "2005 AMC 10B Problems/Problem 25"

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== Solution ==
 
== Solution ==
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[b]Solution 1[/b]
 
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>.
 +
 +
[b]Solution 2[/b]
 +
"Cut" 125 into half. The maximum integer value in the smaller half is 62. 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 12:33, 6 July 2016

Problem

A subset $B$ of the set of integers from $1$ to $100$, inclusive, has the property that no two elements of $B$ sum to $125$. What is the maximum possible number of elements in $B$?

$\mathrm{(A)} 50 \qquad \mathrm{(B)} 51 \qquad \mathrm{(C)} 62 \qquad \mathrm{(D)} 65 \qquad \mathrm{(E)} 68$

Solution

[b]Solution 1[/b] The question asks for the maximum possible number of elements. The integers from $1$ to $24$ can be included because you cannot make $125$ with integers from $1$ to $24$ without the other number being greater than $100$. The integers from $25$ to $100$ are left. They can be paired so the sum is $125$: $25+100$, $26+99$, $27+98$, $\ldots$, $62+63$. That is $38$ pairs, and at most one number from each pair can be included in the set. The total is $24 + 38 = \boxed{\mathrm{(C)}\ 62}$. Also, it is possible to see that since the numbers $1$ to $24$ are in the set there are only the numbers $25$ to $100$ to consider. As $62+63$ gives $125$, the numbers $25$ to $62$ can be put in subset $B$ without having two numbers add up to $125$. In this way, subset $B$ will have the numbers $1$ to $62$, and so $\boxed{\mathrm{(C)}\ 62}$.

[b]Solution 2[/b] "Cut" 125 into half. The maximum integer value in the smaller half is 62. Thus the answer is $\boxed{\mathrm{(C)}\62}$ (Error compiling LaTeX. Unknown error_msg).

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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