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Square Digit Sum Surmise (Posted on 2015-01-20) Difficulty: 3 of 5
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.

No Solution Yet Submitted by K Sengupta    
Rating: 3.0000 (1 votes)

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Solution Challenge accepted (spoiler) | Comment 1 of 2
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
---     --------------
9       45
10     64
18     288

Edited on January 20, 2015, 7:26 pm
  Posted by Steve Herman on 2015-01-20 17:12:54

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