1971 AHSME Problems/Problem 32

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 1

Multiply both sides of the given equation by $(1-2^{-\frac1{32}})$. Using difference of squares reveals that $(1-2^{-\frac1{32}})(1+2^{-\frac1{32}})=(1-2^{-\frac1{16}})$. Thus, we can use difference of squares again on the next term, $(1+2^{-\frac1{16}})$, and the terms after that until we get $(1+2^{-\frac1{32}})s=(1-2^{-\frac12})(1+2^{-\frac12})=1-2^{-1}=\tfrac12$. Dividing both sides by $(1+2^{-\frac1{32}})$ reveals that $s=\boxed{\textbf{(A) }\tfrac12(1+2^{-\frac1{32}})^{-1}}$.

Solution 2 (50-50 guess)

For any two reals $a$ and $x$ with $a>0$, $a^x>0$. Thus, we know that all of the terms $2^{-\frac1x}>0$, so $s$ is the product of $5$ real terms greater than $1$. Thus, $s>1$, so we can eliminate options (C), (D), and (E). Now, we have a $50$% chance of guessing the right answer, $\boxed{\textbf{(A) }\tfrac12(1+2^{-\frac1{32}})^{-1}}$.

See Also

1971 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 31
Followed by
Problem 33
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