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 equal to $5$ real terms multiplied together with each term being 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, $\boxed{\textbf{(A) }\tfrac12(1+2^{-\frac1{32}})^{-1}}$ .

1971 AHSC (Problems • Answer Key • Resources)
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.

Page

Toolbox

Search

1971 AHSME Problems/Problem 32

Contents

Problem

Solution 1

Solution 2 (50-50 guess)

See Also