1962 AHSME Problems/Problem 22: Difference between revisions
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The number <math>121_b</math>, written in the integral base <math>b</math>, is the square of an integer, for | The number <math>121_b</math>, written in the integral base <math>b</math>, is the square of an integer, for | ||
<math> \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 </math> | <math> \textbf{(A)}\ b = 10,\text{ only}\qquad\textbf{(B)}\ b = 10\text{ and }b = 5,\text{ only}\qquad</math> | ||
<math>\textbf{(C)}\ 2\leq b\leq 10\qquad\textbf{(D)}\ b > 2\qquad\textbf{(E)}\ \text{no value of }b </math> | |||
==Solution== | ==Solution== | ||
<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>. | |||
Note that <math>b>2</math> because 2 is a digit in the base-b representation, but for any | |||
<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 
, written in the integral base 
, is the square of an integer, for
Solution
can be represented in base 10 as 
, which factors as 
.
Note that 
because 2 is a digit in the base-b representation, but for any
, is the square of 
. 
