1971 AHSME Problems/Problem 32

Revision as of 16:19, 8 August 2024 by Thepowerful456 (talk | contribs) (added answer statement and see also)

Problem

If $s=(1+2^{-\frac{1}{32}})(1+2^{-\frac{1}{16}})(1+2^{-\frac{1}{8}})(1+2^{-\frac{1}{4}})(1+2^{-\frac{1}{2}})$, then $s$ is equal to

$\textbf{(A) }\textstyle{\frac{1}{2}}(1-2^{-\frac{1}{32}})^{-1}\qquad \textbf{(B) }(1-2^{-\frac{1}{32}})^{-1}\qquad \textbf{(C) }1-2^{-\frac{1}{32}}\qquad \\ \textbf{(D) }\textstyle{\frac{1}{2}}(1-2^{-\frac{1}{32}})\qquad  \textbf{(E) }\frac{1}{2}$


Solution

$\boxed{\textbf{(A) }\tfrac12(1+2^{-\frac1{32}})^{-1}}$.

See Also

1971 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 31
Followed by
Problem 33
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 31 32 33 34 35
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png