First multiple of 2019 with last digits 2018

[Mr Davis 97]

Homework Statement

What positive integer is the first multiple of $2019$ such that the last four digits of this numbers are $2018$?

Homework Equations

The Attempt at a Solution

I was thinking just take multiples of 2019 mod 10000 and look for a pattern, but this didn't really seem to work, so I'm stuck

 

[scottdave]

This is a thought. Set it up like one of those puzzles with the "by hand" multiplication.

Something like this:

...DCBA
   x  2019
----------
...2018

where each letter is a digit.
So then you go 9 times A has to have 8 as the ones digit. The only answer is 18, so A must be 2, and you carry a 1.
Then you have 9 time B plus the 1 carry plus ones digit of (1 times A). Then keep working along those lines.

 

[Mr Davis 97]

This is a thought. Set it up like one of those puzzles with the "by hand" multiplication.

Something like this:

...DCBA
   x  2019
----------
...2018

where each letter is a digit.
So then you go 9 times A has to have 8 as the ones digit. The only answer is 18, so A must be 2, and you carry a 1.
Then you have 9 time B plus the 1 carry plus ones digit of (1 times A). Then keep working along those lines.

I'm a little confused by this. There, you have the last line has having the digit 2018, which is supposed to be the final number. But when doing long multiplication, isn't the case that the first line is only $9$ multiplied by everything, and then we have a second line that is $1$ multiplied by everything, and then we sum these up, and that number would have the last digits as 2018?

 

[scottdave]

I'm a little confused by this. There, you have the last line has having the digit 2018, which is supposed to be the final number. But when doing long multiplication, isn't the case that the first line is only $9$ multiplied by everything, and then we have a second line that is $1$ multiplied by everything, and then we sum these up, and that number would have the last digits as 2018?

Each column is shifted, because the 1 actually represents 1 x 10. The 2 represents 2 x 10³.

 

[haruspex]

So then you go 9 times A has to have 8 as the ones digit. The only answer is 18, so A must be 2,

Good idea, but I suggest the next step would be to subtract 2x2019 from the 2018, borrowing from the left as necessary. Drop the resulting trailing 0 and drop the A, then continue with the B.