Each of a, b, and c is a different base ten digit and n is a positive integer such that:
ab2+c2=ac2+cb2=n
For example:
272+12=212+172=730
Determine the total number of values of the positive integer constant n less than 2023, such that we will have valid values of a,b, and c satisfying the abovementioned condition.
***** No number can contain any leading zero.
(In reply to
Solution by Brian Smith)
34^2=1156, but 34^2+1^2=1157. Similarly, 72^2=5184, but 72^2+1^2=5185. Therefore, the actual answers are 730, 1157, and 1850.
27^2+1^2=21^2+17^2=730
34^2+1^2=31^2+14^2=1157
43^2+1^2=41^2+13^2=1850
In the posted solution, 1156 should be changed to 1157.
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Posted by Math Man
on 2023-05-06 20:35:48 |
re: Solution
