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

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\text{(E)} \text{not uniquely determined} </math> | \text{(E)} \text{not uniquely determined} </math> | ||

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A perfect square will be <math>(8k+r)^2=64k^2+16kr+r^2\equiv r^2\pmod{16}</math> where <math>r=0,1,...,7</math>. | A perfect square will be <math>(8k+r)^2=64k^2+16kr+r^2\equiv r^2\pmod{16}</math> where <math>r=0,1,...,7</math>. | ||

## Revision as of 15:49, 17 June 2021

## Problem 26

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

## 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