N is a positive integer less than 2015.
Find all values of N such that:
N is equal to precisely 36 less than the square of the sum of digits of N.
***As an extra challenge, solve this puzzle without using a computer program assisted method.
The max N is 2014.
This makes the max(sum of digits) = 28 (corresponding to 1999).
28^2 - 36 = 748, so Max N is really 748.
This makes the max(sum of digits) = 24 (corresponding to 699).
24^2 - 36 = 540, so Max N is really 540.
This makes the max(sum of digits) = 22 (corresponding to 499).
22^2 - 36 = 448, so Max N is really 448.
This makes the max(sum of digits) = 21 (corresponding to 399).
21^2 - 36 = 405, so Max N is really 405.
But this doesn't help lower the max sum of digits, which must be greater than 6 and less than or equal to 21.
Squaring all numbers between 7 and 21 and subtracting 36 finds only 3 values of N that work:
SOD N = SOD^2 - 36
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9 45
10 64
18 288
Edited on January 20, 2015, 7:26 pm
Challenge accepted (spoiler)
