Find all possible two digit positive integers N for which the sum of digits of 10^N - N is divisible by 170.

The sum of the digits of 10^N-N is
 
S = 9*(N-2) + {floor(N'/10) + [N' - 10*floor(N'/10)]}

where the term in { } is the sum of the digits of N' = 100 - N.  This may be simplified to

S = 8*(N-1) + 9*ceiling(N/10).

Approximating ceiling(N/10) by N/10 then gives

S ~ 89*N/10 - 8

and equating this to 170*K and solving for N yields

N ~ 10*(170*K + 8)/89 ~ 20, 39.1, 58.2, 77.3, 96.4

for K=1,2,3,4,5. Rounding these off gives the results sought.

Posted by Richard on 2006-07-21 04:05:40