Prove that for every n∈N∗ exists a multiple of n, having sum of digits equal to n.
Problem
Source: Romania JBTST 2008, Problem 2
Tags: number theory proposed, number theory
02.05.2008 00:49
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
04.05.2008 21:48
I saw this problem in the Andrei Negut's book: "Problems for the mathematical Olympiads".
17.07.2008 17:04
Can't we just multiply the number by 10? 10n is a multiple of n and has the same sum of digits.
18.07.2008 13:18
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?
02.01.2009 12:26
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.
03.11.2018 22:10
Can anyone solve this or has it become forgotten?