1982 AHSME Problems/Problem 26
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
See Also
1982 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 25 |
Followed by Problem 27 | |
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 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.