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 $9$ .

#NumberTheory

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Comments

Can you give the numbers? @ChengYiin Ong

A Former Brilliant Member - 1 year 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 $9$ .

ChengYiin Ong - 1 year 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?"

Sahar Bano - 1 year 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"

ChengYiin Ong - 1 year ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$