# A seemingly hard divisibility problem

For every $n=1,2,3,..., 2040$, prove there exists a multiple of $n$, which has less than $12$ digits, and only contains the digits $0,1,8$ or $9$.

Solution: Take two $11$ digits numbers, consisting of only $0$ and $1$, since there are $2^{11}=2048$ numbers that can be formed, there must be two numbers with the same remainder after dividing by $n$. Thus, subtracting one with another, we get a multiple of $n$ which has less than $12$ digits, and only contains the digits $0,1,8$ or $9$.

Generalization: For every $n=1,2,3,..., k$ where $k \le 2^j-1$, there exists a multiple of $n$, which has less than $j+1$ digits, and only contains the digits $0,1,8$ or $9$.

Note by ChengYiin Ong
5 months, 3 weeks ago

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Can you give the numbers? @ChengYiin Ong

- 5 months, 3 weeks ago

Maybe for $n=11$, i can choose $11100001001$ and $10000000000$, they both have $11$ digits and subtracting the first with the second, we get $1100001001$ which is a multiple of 11 and only contains the digits $0,1,8$ or $9$.

- 5 months, 3 weeks ago

@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?"

- 5 months, 3 weeks ago

there are $2048$ numbers that you can form from choosing only $0$ or $1$ at for every digit and so there must be two numbers that have the same modulo $n$, maybe i shouldn't say "possibility"

- 5 months, 3 weeks ago