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

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<math>\textbf{(A) } 50 \qquad \textbf{(B) } 51 \qquad \textbf{(C) } 62 \qquad \textbf{(D) } 65 \qquad \textbf{(E) } 68 </math>
 
<math>\textbf{(A) } 50 \qquad \textbf{(B) } 51 \qquad \textbf{(C) } 62 \qquad \textbf{(D) } 65 \qquad \textbf{(E) } 68 </math>
  
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==-Solutions-==
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==Solution 1==
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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{\textbf{(C)}\ 62}</math>.
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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 the answer is <math>\boxed{\mathrm{(C)}\ 62}</math>.
 
  
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====Solution 1 Alternate Solution====
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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 the answer is <math>\boxed{\textbf{(C)}\ 62}</math>.
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Since there are 38 numbers that sum to <math>125</math>, there are <math>100-38=62</math> numbers not summing to <math>125.</math>
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===Solution 1 Alternate Solution===
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Since there are <math>38</math> numbers that sum to <math>125</math>, there are <math>100-38=62</math> numbers not summing to <math>125.</math>
 
~mathboy282
 
~mathboy282
  
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===Solution 2 (If you have no time)===
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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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"Cut" <math>125</math> into half. The maximum integer value in the smaller half is <math>62</math>. Thus the answer is <math>\boxed{\textbf{(C)}\ 62}</math>.
  
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===Solution 3===
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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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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 =\boxed{\textbf{(C)}\ 62}</math>.
  
 
~GentleTiger
 
~GentleTiger

Revision as of 17:17, 16 December 2021

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 $B$?

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

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 $24 + 38 = \boxed{\textbf{(C)}\ 62}$.

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 the answer is $\boxed{\textbf{(C)}\ 62}$.

Solution 1 Alternate Solution

Since there are $38$ numbers that sum to $125$, there are $100-38=62$ numbers not summing to $125.$ ~mathboy282

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

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 $n =\boxed{\textbf{(C)}\ 62}$.

~GentleTiger

See Also

2005 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Problem
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All AMC 10 Problems and Solutions

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