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 | ||
− | < | + | <cmath>\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}+\frac{1}{12}</cmath> |
− | if the sum of the remaining terms is to equal <math> 1 </math>? | + | 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== | ||
− | + | 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> | |
+ | 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>. | ||
− | <math> | + | 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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==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
if the sum of the remaining terms is to equal ?
Solution
We compute the entire sum as so the two terms to be removed must sum to . That is, for some .
Since and are coprime, we deduce that is a multiple of , so or must be a multiple of . It follows that one of them must be , and we obtain so the two terms we must remove are .
See Also
1985 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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All AHSME Problems and Solutions |
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