Difference between revisions of "1962 AHSME Problems/Problem 22"

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==Solution==
 
==Solution==
{{solution}}
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<math>121_b</math> can be represented in base 10 as <math>b^2+2b+1</math>, which factors as <math>(b+1)^2</math>.
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Note that <math>b>2</math> because 2 is a digit in the base-b representation, but for any
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<math>b>2</math>, <math>121_b</math> is the square of <math>b+1</math>. <math>\boxed{\textbf{(D)}}</math>

Latest revision as of 15:27, 16 April 2014

Problem

The number $121_b$, written in the integral base $b$, is the square of an integer, for

$\textbf{(A)}\ b = 10,\text{ only}\qquad\textbf{(B)}\ b = 10\text{ and }b = 5,\text{ only}\qquad$

$\textbf{(C)}\ 2\leq b\leq 10\qquad\textbf{(D)}\ b > 2\qquad\textbf{(E)}\ \text{no value of }b$

Solution

$121_b$ can be represented in base 10 as $b^2+2b+1$, which factors as $(b+1)^2$. Note that $b>2$ because 2 is a digit in the base-b representation, but for any $b>2$, $121_b$ is the square of $b+1$. $\boxed{\textbf{(D)}}$