Positive integers such that the sum of its digits is a third of itself

Positive integers such that the sum of their digits is a third of itself

1 DIGIT:

3a=aa=3a

The only 1 digit integer that satisfies this is 0, but 0∉Z+so there is no solution.

2 DIGITS:

3ab=a+b10a+b=3a+3b7a=2ba∣2∧b∣7∧a,b≤9⇒a=2∧b=7

Therefore, the only 2 digit integer that satisfies this is 27.

3 DIGITS:

3abc=a+b+c100a+10b+c=3a+3b+3c97a+7b=2c

For abc to be a 3 digit integer, a=0. If a=1, it's alrrady impossible for the equation to hold true, as 97+7b=2c, and because b≥0, we're saying that 2c≥97, which is impossible for values of c≤9. Therefore, there are no 3 digit integers that satisfy this property. Similarly, with 4 digit integers:

1000a+100b+10c+d=3a+3b+3c+3d997a+97b+7c=2d

Again, if (0<a≤9)∧(0≤b,c,d≤9)there are no values of 2d≥997. Therefore, there are no 4 or greater digit integers that satisfy this property

In conclusion, the only positive integer such that the sum of their digits is a third of itself is 27. Q.E.D.