1979 AHSME Problems/Problem 11

Problem 11

Find a positive integral solution to the equation $\frac{1+3+5+\dots+(2n-1)}{2+4+6+\dots+2n}=\frac{115}{116}$

$\textbf{(A) }110\qquad \textbf{(B) }115\qquad \textbf{(C) }116\qquad \textbf{(D) }231\qquad\\ \textbf{(E) }\text{The equation has no positive integral solutions.}$

Solution

Solution by e_power_pi_times_i

Notice that the numerator and denominator are the sum of the first $n$ odd and even numbers, respectively. Then the numerator is $n^2$ , and the denominator is $n(n+1)$ . Then $\frac{n}{n+1} = \frac{115}{116}$ , so $n = \boxed{\textbf{(B) } 115}$ .

1979 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 10
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1979 AHSME Problems/Problem 11

Problem 11

Solution

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