A seemingly hard divisibility problem

For every n=1,2,3,...,2040n=1,2,3,..., 2040, prove there exists a multiple of nn, which has less than 1212 digits, and only contains the digits 0,1,80,1,8 or 99.

Solution: Take two 1111 digits numbers, consisting of only 00 and 11, since there are 211=20482^{11}=2048 numbers that can be formed, there must be two numbers with the same remainder after dividing by nn. Thus, subtracting one with another, we get a multiple of nn which has less than 1212 digits, and only contains the digits 0,1,80,1,8 or 99.

Generalization: For every n=1,2,3,...,kn=1,2,3,..., k where k2j1k \le 2^j-1, there exists a multiple of nn, which has less than j+1j+1 digits, and only contains the digits 0,1,80,1,8 or 99.

Note by ChengYiin Ong
3 weeks ago

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Can you give the numbers? @ChengYiin Ong

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Maybe for n=11n=11, i can choose 1110000100111100001001 and 1000000000010000000000, they both have 1111 digits and subtracting the first with the second, we get 11000010011100001001 which is a multiple of 11 and only contains the digits 0,1,80,1,8 or 99.

ChengYiin Ong - 3 weeks ago

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@ChengYiin Ong - if there are 2048 possibilities then how can you say that "there must be two numbers with the same remainder after dividing by n?"

Sahar Bano - 3 weeks ago

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there are 20482048 numbers that you can form from choosing only 00 or 11 at for every digit and so there must be two numbers that have the same modulo nn, maybe i shouldn't say "possibility"

ChengYiin Ong - 3 weeks ago

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