Problem of the week 2 -

[Math Tutor]

Is it possible to find a natural number [tex]k,[/tex] so that [tex]S(k)+S(k^2)=2009,[/tex] where [tex]S(k)[/tex] is the sum of the digits of [tex]k[/tex].

[bigmamma]

Yes. The number 999....91, where the digit 9 appeared 111 times, satisfies the conditon. It is easy to check.

[Math Tutor]

The answer is correct.
How did you solve it?

[bigmamma]

999........91, where 9 appears n times and its square differ with 9 in the sum of the digits. 999...91=10[sup]n+1[/sup]-9 has sum of the digits n*9+1. Its square
(10[sup]n+1[/sup]-9)[sup]2[/sup]=10[sup]2n+2[/sup]-18*10[sup]n+1[/sup]+81 has sum of digits 9*n +10. Since 2009=1000+1009 we obtain the solution.