Question 8
A 5-digit number (in base 10) has digits $$k, k+1, k+2, 3k, k+3$$ in that order, from left to right. If this number is $$m^2$$ for some natural number $$m$$, find the sum of the digits of $$m$$.
Correct Answer: 15
Solution
All five entries must be single digits, so $$3k \le 9$$ forces $$k \le 3$$, giving the candidates 12334, 23465 and 34596. Only the last is a perfect square, since $$186^2 = 34596$$. Hence $$m = 186$$ and the sum of its digits is $$1+8+6 = 15$$.
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