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Difference between revisions of "1985 AHSME Problems/Problem 5"

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Which terms must be removed from the sum
 
Which terms must be removed from the sum
  
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<math> \frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}+\frac{1}{12} </math>
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<cmath>\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}+\frac{1}{12}</cmath>
  
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if the sum of the remaining terms is to equal <math> 1 </math>?
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if the sum of the remaining terms is to equal <math>1</math>?
  
 
<math> \mathrm{(A)\ } \frac{1}{4}\text{ and }\frac{1}{8} \qquad \mathrm{(B) \ }\frac{1}{4}\text{ and }\frac{1}{12} \qquad \mathrm{(C) \  } \frac{1}{8}\text{ and }\frac{1}{12} \qquad \mathrm{(D) \  } \frac{1}{6}\text{ and }\frac{1}{10} \qquad \mathrm{(E) \  }\frac{1}{8}\text{ and }\frac{1}{10}  </math>
 
<math> \mathrm{(A)\ } \frac{1}{4}\text{ and }\frac{1}{8} \qquad \mathrm{(B) \ }\frac{1}{4}\text{ and }\frac{1}{12} \qquad \mathrm{(C) \  } \frac{1}{8}\text{ and }\frac{1}{12} \qquad \mathrm{(D) \  } \frac{1}{6}\text{ and }\frac{1}{10} \qquad \mathrm{(E) \  }\frac{1}{8}\text{ and }\frac{1}{10}  </math>
  
 
==Solution==
 
==Solution==
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First, we sum all of the terms to see how much more than <math> 1 </math> the entire sum is.
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We compute the entire sum as <cmath>\begin{align*}\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}+\frac{1}{12} &= \frac{60+30+20+15+12+10}{120} \\ &= \frac{147}{120} \\ &= \frac{49}{40},\end{align*}</cmath>
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so the two terms to be removed must sum to <math>\frac{49}{40}-1 = \frac{9}{40}</math>. That is, <math>\frac{9}{40} = \frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}</math> for some <math>x,y \in \{2,4,6,8,10,12\}</math>.
  
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<math> \frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}+\frac{1}{12} </math>
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Since <math>9</math> and <math>40</math> are coprime, we deduce that <math>xy</math> is a multiple of <math>40</math>, so <math>x</math> or <math>y</math> must be a multiple of <math>5</math>. It follows that one of them must be <math>10</math>, and we obtain <cmath>\begin{align*}\frac{9}{40}-\frac{1}{10} &= \frac{9-4}{40} \\ &= \frac{5}{40} \\ &= \frac{1}{8},\end{align*}</cmath> so the two terms we must remove are <math>\boxed{\text{(E)} \ \frac{1}{8}\text{ and }\frac{1}{10}}</math>.
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<math> \frac{60}{120}+\frac{30}{120}+\frac{20}{120}+\frac{15}{120}+\frac{12}{120}+\frac{10}{120} </math>
 
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<math> \frac{147}{120} </math>
 
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<math> \frac{49}{40} </math>
 
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So we want two of the terms that sum to <math> \frac{49}{40}-1=\frac{9}{40} </math>.
 
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Consider <math> \frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}=\frac{9}{40} </math>. Therefore, we must have <math> 40 </math> as a factor of <math> xy </math>. Notice that <math> 10 </math> is the only possible value of <math> x </math> and <math> y </math> that's a multiple of <math> 5 </math>, so one of <math> x </math> or <math> y </math> must be <math> 10 </math>. Now subtract <math> \frac{9}{40}-\frac{1}{10}=\frac{9}{40}-\frac{4}{40}=\frac{5}{40}=\frac{1}{8} </math>, so the two fractions we must remove are <math> \frac{1}{8}\text{ and }\frac{1}{10}, \boxed{\text{E}} </math>.
 
  
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1985|num-b=4|num-a=6}}
 
{{AHSME box|year=1985|num-b=4|num-a=6}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 17:55, 19 March 2024

Problem

Which terms must be removed from the sum

\[\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}+\frac{1}{12}\]

if the sum of the remaining terms is to equal $1$?

$\mathrm{(A)\ } \frac{1}{4}\text{ and }\frac{1}{8} \qquad \mathrm{(B) \ }\frac{1}{4}\text{ and }\frac{1}{12} \qquad \mathrm{(C) \ } \frac{1}{8}\text{ and }\frac{1}{12} \qquad \mathrm{(D) \ } \frac{1}{6}\text{ and }\frac{1}{10} \qquad \mathrm{(E) \ }\frac{1}{8}\text{ and }\frac{1}{10}$

Solution

We compute the entire sum as \begin{align*}\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}+\frac{1}{12} &= \frac{60+30+20+15+12+10}{120} \\ &= \frac{147}{120} \\ &= \frac{49}{40},\end{align*} so the two terms to be removed must sum to $\frac{49}{40}-1 = \frac{9}{40}$. That is, $\frac{9}{40} = \frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}$ for some $x,y \in \{2,4,6,8,10,12\}$.

Since $9$ and $40$ are coprime, we deduce that $xy$ is a multiple of $40$, so $x$ or $y$ must be a multiple of $5$. It follows that one of them must be $10$, and we obtain \begin{align*}\frac{9}{40}-\frac{1}{10} &= \frac{9-4}{40} \\ &= \frac{5}{40} \\ &= \frac{1}{8},\end{align*} so the two terms we must remove are $\boxed{\text{(E)} \ \frac{1}{8}\text{ and }\frac{1}{10}}$.

See Also

1985 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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