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

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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>
 
<math>\mathrm{(A)} 50 \qquad \mathrm{(B)} 51 \qquad \mathrm{(C)} 62 \qquad \mathrm{(D)} 65 \qquad \mathrm{(E)} 68 </math>
  
==Solutions==
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==-Solutions-==
 
===Solution 1===
 
===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>.

Revision as of 12:50, 10 January 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 $B$?

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

-Solutions-

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

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

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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