...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,
To find the first multiple of 2019 with last digits 2018, we need to divide 2018 by 2019. The remainder will give us the last two digits of the multiple. We then subtract the remainder from 2018 to get the first two digits of the multiple. Therefore, the first multiple of 2019 with last digits 2018 is 2018 - (2018 % 2019) = 2018 - 2018 = 0.
The significance of finding the first multiple of 2019 with last digits 2018 is that it can help us understand the patterns and properties of numbers. It also has practical applications in cryptography and number theory.
No, there can only be one first multiple of 2019 with last digits 2018. This is because every number has a unique multiple, and since 2019 is a prime number, it has only two factors (1 and itself), making it impossible for there to be more than one multiple with the same last digits.
The concept of finding the first multiple of 2019 with last digits 2018 is closely related to modular arithmetic. In modular arithmetic, we find the remainder when a number is divided by another number. In this case, we are finding the remainder when 2018 is divided by 2019, which gives us the last two digits of the multiple.
Yes, this concept can be applied to any number. We can find the first multiple of any number with any given last digits by following the same process of dividing and subtracting the remainder. However, the result may not always be a whole number, as some numbers do not have a multiple with the given last digits.