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Difference between revisions of "1971 AHSME Problems/Problem 32"

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== Problem 32 ==
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== Problem ==
  
 
If <math>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}})</math>, then <math>s</math> is equal to
 
If <math>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}})</math>, then <math>s</math> is equal to
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<math>\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}</math>
 
<math>\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}</math>
  
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== Solution ==
 
  
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First, we write the expression as <math>(1+\frac{1}{\sqrt[32]{2}})(1+\frac{1}{\sqrt[16]{2}})(1+\frac{1}{\sqrt[8]{2}})(1+\frac{1}{\sqrt[4]{2}})(1+\frac{1}{\sqrt[2]{2}})</math>, or <math>(1+\frac{1}{\sqrt[32]{2^1}})(1+\frac{1}{\sqrt[32]{2^2}})(1+\frac{1}{\sqrt[32]{2^4}})(1+\frac{1}{\sqrt[32]{2^8}})(1+\frac{1}{\sqrt[32]{2^{16}}})</math>
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== Solution 1 ==
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Multiply both sides of the given equation by <math>(1-2^{-\frac1{32}})</math>. Using [[difference of squares]] reveals that <math>(1-2^{-\frac1{32}})(1+2^{-\frac1{32}})=(1-2^{-\frac1{16}})</math>. Thus, we can use difference of squares again on the next term, <math>(1+2^{-\frac1{16}})</math>, and the terms after that until we get <math>(1+2^{-\frac1{32}})s=(1-2^{-\frac12})(1+2^{-\frac12})=1-2^{-1}=\tfrac12</math>. Dividing both sides by <math>(1+2^{-\frac1{32}})</math> reveals that <math>s=\boxed{\textbf{(A) }\tfrac12(1+2^{-\frac1{32}})^{-1}}</math>.
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== Solution 2 (50-50 guess) ==
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For any two reals <math>a</math> and <math>x</math> with <math>a>0</math>, <math>a^x>0</math>. Thus, we know that all of the terms <math>2^{-\frac1x}>0</math>, so <math>s</math> is the product of <math>5</math> real terms greater than <math>1</math>. Thus, <math>s>1</math>, so we can eliminate options (C), (D), and (E). Now, we have a <math>50</math>% chance of guessing the right answer, <math>\boxed{\textbf{(A) }\tfrac12(1+2^{-\frac1{32}})^{-1}}</math>.
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== See Also ==
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{{AHSME 35p box|year=1971|num-b=31|num-a=33}}
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{{MAA Notice}}
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[[Category:Intermediate Algebra Problems]]

Latest revision as of 16:36, 8 August 2024

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 (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

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