1979 AHSME Problems/Problem 6

Problem 6

$\frac{3}{2}+\frac{5}{4}+\frac{9}{8}+\frac{17}{16}+\frac{33}{32}+\frac{65}{64}-7=$

$\textbf{(A) }-\frac{1}{64}\qquad \textbf{(B) }-\frac{1}{16}\qquad \textbf{(C) }0\qquad \textbf{(D) }\frac{1}{16}\qquad \textbf{(E) }\frac{1}{64}$

Solution

Solution by e_power_pi_times_i

Simplifying, we have $\frac{96}{64}+\frac{80}{64}+\frac{72}{64}+\frac{68}{64}+\frac{66}{64}+\frac{65}{64}-7$ , which is $\frac{96+80+72+68+66+65}{64}-7 = \frac{447}{64}-7 = 6\frac{63}{64}-7 = \boxed{\textbf{(A) } -\frac{1}{64}}$ .

1991 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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1979 AHSME Problems/Problem 6

Problem 6

Solution

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