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Difference between revisions of "1979 AHSME Problems/Problem 6"

(Created page with "== Problem 6 == <math>\frac{3}{2}+\frac{5}{4}+\frac{9}{8}+\frac{17}{16}+\frac{33}{32}+\frac{65}{64}-7=</math> <math>\textbf{(A) }-\frac{1}{64}\qquad \textbf{(B) }-\frac{1}{...")
 
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== See also ==
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{{AHSME box|year=1991|num-b=22|num-a=24}}   
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{{AHSME box|year=1979|num-b=22|num-a=24}}   
  
 
[[Category: Introductory Algebra Problems]]
 
[[Category: Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:01, 5 January 2017

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

See also

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