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

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== Problem ==
 
== Problem ==
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== Solution ==
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A subset <math>B</math> of the set of integers from <math>1</math> to <math>100</math>, inclusive, has the property that no two elements of <math>B</math> sum to <math>125</math>. What is the maximum possible number of elements in <math>B</math>?
  
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The question asks for the maximum possible. The integers from 1~24 can be included because you cannot make 125 with integers from 1~24 without the other number being greater than 100. The integers 25~100 are left. They can be paired so the sum is 125. 25+100, 26+99, 27+98, ...... 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 = 62 --> C.
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<math>\mathrm{(A)} 50 \qquad \mathrm{(B)} 51 \qquad \mathrm{(C)} 62 \qquad \mathrm{(D)} 65 \qquad \mathrm{(E)} 68 </math>
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==Solution==
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===Solution 1===
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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>.
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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>.
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===Solution 2 (If you have no time)===
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"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>.
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===Solution 3===
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The maximum possible number of elements includes the smallest numbers. So, subset <math>B = \{1,2,3....n-1,n\}</math>  where n is the maximum number of elements in subset <math>B</math>. So, we have to find two consecutive numbers, <math>n</math> and <math>n+1</math>, whose sum is <math>125</math>. Setting up our equation, we have <math>n+(n+1) = 2n+1 = 125</math>. When we solve for <math>n</math>, we get <math>n = 62</math>. Thus, the anser is <math>\boxed{\mathrm{(C)}\ 62}</math>.
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~GentleTiger
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lol this should have been like #13 or so
  
 
== See Also ==
 
== See Also ==
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*[[2005 AMC 10B Problems]]
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{{AMC10 box|year=2005|ab=B|num-b=24|after=Last Problem}}
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{{MAA Notice}}

Revision as of 14:09, 22 September 2020

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

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

Solution 3

The maximum possible number of elements includes the smallest numbers. So, subset $B = \{1,2,3....n-1,n\}$ where n is the maximum number of elements in subset $B$. So, we have to find two consecutive numbers, $n$ and $n+1$, whose sum is $125$. Setting up our equation, we have $n+(n+1) = 2n+1 = 125$. When we solve for $n$, we get $n = 62$. Thus, the anser is $\boxed{\mathrm{(C)}\ 62}$.

~GentleTiger

lol this should have been like #13 or so

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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