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^NN is
S = 9*(N2) + {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*(N1) + 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 20060721 04:05:40 