Find all integers n such that S(n) = S(2n) = ... = S(n*n), where S is the sum of the base-10 digits.
(In reply to
The first answer (spoiler) by Steve Herman)
You say
////Arguably 0 and 1 work also, but I think the form of the problem implies that N >=2.
/////
0 -yes 1 - no
The answer is 9, 99, 999 ,,, 10^k-1 k=0,1,2,3...
The general proof is tiresome and it involves to many indices , so I will just show why it works for 99 multiples
0f 99:
(10a+b)(99)=(10a+b)(100-1)=1000a+100(b-1)+10(9-a)+10-b
the sod of the right side is a+b-1+9-a+10-b=-1+9+10=18 like sod 0f 99.
e.g. 99*64=6400-64=6336, 6+3+3+6=18
Edited on November 22, 2012, 4:46 pm
re: The first answer (spoiler)
