The problem involves finding the first positive integer that is a multiple of 2019, with the last four digits being 2018. This falls under the subject area of number theory and modular arithmetic.
The discussion is ongoing, with participants exploring different methods to approach the problem. Some guidance has been offered regarding setting up the multiplication and considering the implications of carrying digits. There is a recognition of confusion around the multiplication setup and how it relates to the final digits.
Participants are questioning the assumptions made about the multiplication process and the representation of digits in the context of the problem. There is an acknowledgment of the complexity involved in handling carries and the structure of the multiplication.
...DCBA
x 2019
----------
...2018I'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 said:This is a thought. Set it up like one of those puzzles with the "by hand" multiplication.
Something like this:
where each letter is a digit.Code:...DCBA x 2019 ---------- ...2018
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.
Each column is shifted, because the 1 actually represents 1 x 10. The 2 represents 2 x 10³.Mr Davis 97 said: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?
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.scottdave said: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,