decimal expansion of an integer

Question

Can someone be so kind as to explain what is meant by the decimal expansion of an integer? I saw the following at this link but I don't know what decimal expansion of an integer refers to: https://madhavamathcompetition.com/2017/01/23/the-pigeon-hole-principle-some-notes-and-examples/ Example 4:

Show that for every integer $n$ there is a multiple of $n$ that has only $0$’s and $1$’s in its decimal expansion.

Solution 4:

Let $n$ be a positive integer. Consider the $n+1$ integers $1, 11, 111, \ldots, 11\ldots 1$ (where the last integer in this list is the integer with $n+1$ $1$’s in its decimal expansion). Note that there are $n$ possible remainders when an integer is divided by $n$. Because there are $n+1$ integers in this list, by the pigeonhole principle, there must be two with the same remainder when divided by $n$. The larger of these integers less the smaller one is a multiple of $n$, which has a decimal expansion consisting entirely of $0$’s and $1$’s.

Perhaps I am misunderstanding the use of the word multiple. I thought it would mean an integer multiple.

Answer 1

Perhaps it helps to do a simple example. Take $n=6$. We have:

$1=6\times 0+1$

$11=6\times 1+5$

$111=6\times 18 +3$

$1111=6\times 185 + 1$

$11111=6\times 1851+5$

$111111=6\times 18518 +3$

$1111111=6\times 185185 +1$

There are seven numbers here, and only six possible remainders, so two of the remainders must be the same. We pick the lowest two and take their difference:

$1111-1=1110=185\times 6$

and we have an integer multiple of $6$ which consists (in decimal notation) of a string of $0$s and $1$s.

The proof works with strings of $1$s to get the remainders and then takes one string from another, which leaves some trailing zeros. You might like to think about how many zeros you expect (it has to do with how the factors $2$ and $5$ appear in $n$).