Let n= (2^a). (5^b). L where L is not divisible by 2 or 5. Let k be a positive integer such that L|10^k -1 Take N= 10^(a+b)* [10^k + 10^(k^2) +10^(k^3)+....+ 10^(k^n) ] We have n|N and S(N) =n

I saw this problem in the Andrei Negut's book: "Problems for the mathematical Olympiads".

Can't we just multiply the number by 10? $10n$ is a multiple of $n$ and has the same sum of digits.

you need the sum of digits to be equal to n. not the sum of digits to be equal to the sum of digits of n. for example when n=12 you need a multiple that is divisible by 12 and with sum of digits=12 10*12=120 but the sum of digits of 120 is 3. instead a proper multiple in this case would be 444. Do you understand?

I have an interesting observation about this task, 1*1=1 2*1=2 3*1=3 4*1=4 5*1=5 6*19=114 7*19=133 8*19=152 9*19=171 10*19=190 11*19=209 12*19=228 13*19=247 14*19=266 15*19=285 16*199=3184 25*199=4975 26*1999=51974 . . . I have very long and clumsy proof for this so can somebody else try proving it.

Can anyone solve this or has it become forgotten?