# Difference between revisions of "1982 AHSME Problems/Problem 26"

Chessmaster3 (talk | contribs) (→A Solution) |
Chessmaster3 (talk | contribs) (→A Solution) |
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Notice that <math>r^2\equiv 1,4,9,0 \pmod{16}</math>. | Notice that <math>r^2\equiv 1,4,9,0 \pmod{16}</math>. | ||

− | Now <math>ab3c</math> in base 8 is <math>a8^3+b8^2+3(8)+c\equiv 8+c\pmod{16}</math>. It being a perfect square means <math>8+c\equiv 1,4,9,0 \pmod{16}</math>. That means that c can only be 1 so the answer is 1 = <math>\boxed{\textbf{( | + | Now <math>ab3c</math> in base 8 is <math>a8^3+b8^2+3(8)+c\equiv 8+c\pmod{16}</math>. It being a perfect square means <math>8+c\equiv 1,4,9,0 \pmod{16}</math>. That means that c can only be 1 so the answer is 1 = <math>\boxed{\textbf{(B)}.}</math>. |

== Partial and Wrong Solution == | == Partial and Wrong Solution == |

## Revision as of 14:12, 26 January 2019

## Problem 26

If the base representation of a perfect square is , where , then equals

## A Solution

A perfect square will be where .

Notice that .

Now in base 8 is . It being a perfect square means . That means that c can only be 1 so the answer is 1 = .

## Partial and Wrong Solution

From the definition of bases we have , and

If , then , which makes

If , then , which clearly can only have the solution , for . This makes , which doesn't have 4 digits in base 8

If , then , which clearly can only have the solution , for . is greater than , and thus, this solution is invalid.

If , then , which clearly has no solutions for .

Similarly, yields no solutions

If , then , which clearly can only have the solution , for . This makes , which doesn't have 4 digits in base 8.

If , then , which clearly can only have the solution , for . This makes , which doesn't have 4 digits in base 8