Question

How many numbers of (abc), where each letter represents a digit, that (2018abc) is divisible by 3, 7, 13?

Answer 1

If `a != 0` then three, namely `289`, `562` and `835`.

If `a = 0` is permitted then `016` is the fourth possible value.

Explanation:

Since `3`, `7` and `13` are all coprime, their least common multiple is:

`3 * 7 * 13 = 273`

We find:

`2018000 / 273 = 7391 \` with remainder `\ 257`

So the lowest number of the form `2018"abc"` that is divisible by `273` is:

`2018000 + (273-257) = 2018016`

There are three more numbers of this form also divisible by `273` and hence by `3`, `7` and `13`, namely:

`2018016 + 273 = 2018289`

`2018289 + 273 = 2018562`

`2018562 + 273 = 2018835`

It is not clear from the question whether the first digit needs to be non-zero, thus making `"abc"` a three digit number, but if so then that means there are three suitable `3` digit numbers, namely `289`, `562` and `835`.