2021/081) 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.

Because there is digit 3k so k <= 3

takiing k = 1 we have last 2 digits = 34 so it cannot be a square

taking k = 2 last 2 digits are 65 so it cannot be a square

taking k = 3 we get $34596= 186^2$ or m = 186 so sum of digits = 1 + 8 + 6 = 15