Can A Multiple Of 2013 Have These Last Four Digits?

As $gcd ( 2013,10000 ) = 1$ , we see, there are $n,m \in \mathbb{Z}$ such that,

$2013 n + 10000 m = 1$

Thus, $(2013)(2014n) - 2014 = - 2014 (10000 m)$

This implies that, $(2013)(2014n) \equiv 2014 \; (mod \; 10000)$ .

If $n < 0$ , substitute $(2013)(2014n)$ with a positive $(2013)(2014n + 10000 k)$ , where $k$ is an arbitrary integer.

Similarly, we can demonstrate the proof of the last four digits being $2015$ . $_ \square$

We have to find out if there's a number $x \equiv 0 \mod 2013\,, x \equiv \left\lbrace{2014,2015}\right\rbrace \mod 10^4\,.$

Since 2013 and 10000 are coprime, we know that such a number always exists from the Chinese Remainder Theorem.

Fun question: Can you tell me all the numbers which when multiplied to $2013$ end in $2014/2015$ ? An elementary approach solves the posed as well as the above problem. Go back to childhood riddles!

$2 0 1 3$

$xyz\times$

_2014____

(... $x,y,z$ are digits) You know that $z$ can only be $8$ , because only then the last digit of product will be $4$ . There is no other option for $z$ .

$2 0 1 3$

$xy8\times$

=====

$16104$

$????\times$

_2014____

Now among the question marks, the last one must be $1$ , so the multiplication of digit $y$ with $2013$ ends in $1$ , so $y$ is $7$ .

$2 0 1 3$

$x78\times$

=====

_ $16104$

$14091\times$

$????\times\times$

_ _2014__

Obviously, the last question mark now should be $0$ . So $x$ is $0$ .

_ $2 0 1 3$

$w078\times$

=====

_ $16104$

$14091\times$

$0000\times\times$

$???\times\times\times$

_ _2014__

The last question marked digit should now be $5$ . Because, in that column: $6$ and the carry over of $1$ from the right column has already made up a $7$ . So the digit $5$ can only make up the $2$ required in the product. So, digit w multiplied by $2013$ ends in $5$ . Thus, digit $w$ is $5$ . There is no ambiguity.

_ $2 0 1 3$

$5078\times$

=====

_''''" $16104$

_ '' $14091\times$

_ '' $0000\times\times$

$10065\times\times\times$

_ _2014__

So, the last digits, WITHOUT ambiguity are known to be $2014$ if and only if the number to be multiplied ends in $...5078$ . Thus the complete set of numbers can be represented as:

$n\times1000 + 5078$

Of course, the question of possibility is solved easily. Yes, it is possible for a multiple of $2013$ to end in $2014$ . (precisely, $????...??5078$ multiples actually)

A similiar analysis for $2015$ yields a 'yes' for $????...??0155$ multiples.

Interestingly, the procedure makes it clear:

It is possible for numbers ending in any segment of digits to come up as multiples of any odd number for which member of a unique family represented by: $n\times10^p + x$ , must be multiplied ; here $p$ is the segment length (number of digits) and segment $x$ has $p$ digits.

As 2013 x 5077 = 10220001, any ABCD can be the last four digits as a multiple when multiplied by 5077 x ABCD, including 2014 and 2015.

2014 and 2015. According to Dirichlet theory, in sequence: 2014,20142014,......., 20142014....2014 ( 2014 times of 2014), there are 2 numbers which are 20142014..2014 (m times) and 20142014...2014( n times) and the subtraction of them is multiple of 2013 ===> 20142014...2014 (m times) - 20142014...2014 (n times) multiple for 2013 ===> 20142014...2014 (m -n times) is multiple of 2013

OR

5078 2013=10222014, 155 2013=312015, So both 2014 and 2015