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

m (Solution)
m (Solution 2)
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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==
+
==Solution 2 (If you have no time)==
 
"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>.
 
"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>.
  

Revision as of 14:51, 25 November 2018

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 1

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 $\boxed{\mathrm{(C)}\ 62}$.

Solution 2 (If you have no time)

"Cut" $125$ into half. The maximum integer value in the smaller half is $62$. Thus the answer is $\boxed{\mathrm{(C)}\ 62}$.

See Also

2005 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
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All AMC 10 Problems and Solutions

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