Reverse years

See guys, looking at it you can see that it's a linear Diophantine equation in 2 variables. of the form $ax+by=N$ . Now suppose $gcd( a, b)=d$ ,where $a,b,x,y,d$ are integers .Now for the existence of a solution ,as $d$ divides $ax+by$ ..........so in turn, d must divide N.
( Well this is because if $gcd(a,b)=d$ , then d divides both a and b, so d divides $ax+by$ for integers $x,y$ )

So, in order for the existence of solution of the given equation, $gcd(2013,3102)=33$ must divide $N$ . So 33 must divide N,i.e N must be a multiple of 33. Clearly the smallest such N is 33 . Well this is my first time I wrote a solution and also that using the code, so sorry for the disalignment in writing.

We'll try to see a more general case -

By the Euclidean algorithm, we know that $gcd(a,b) = ax + by$

We know that $gcd(a,b)$ gives us the smallest value of $ax+by$ for any integers $x$ and $y$

Hence, in particular the smallest value of $N = 2013x+3102y$ would definetely be $gcd(2013,3102)$

Now $2013 = 3\times11\times61$ and $3102 = 2\times3\times11\times47$ , so $gcd(2013,3102)$ turns out be $33$

Hence, the smallest value of $N = \boxed{33}$

According to Bezout's Lemma, the minimum positive value that can be written in the given form has to be the GCD of the integers (2013, 3012) . GCD of the numbers is 33 . Therefore the solution is 33 .

By Bezout's lemma the smallest possible positive N is gcd(2013, 3102)=33.

Well, the gcd between $2013$ and $3102$ is $33$ , therefore $33 | N$ , i.e.,

$33 \times k = N > 0 \therefore k = 1 \Rightarrow N = 33$

x=38 and y=(-24)

2013x38+3102x(-24)=74484-74481=33

we know that gcd(a,b) can be written as ax+by for integers x,y hence 2013x+3102y = gcd(2013,3102) which is the least N..

From theorem we get, $gcd(a,b)=ax+by$ for any integer x and y.

In the above problem, $2013x+3102y=N$ can be written as, $gcd(2013,3102)=N$ .

By further simple calculation it can be calculated that $gcd(2013,3102)=33$

Therefore The answer is $N=\boxed{33}$ .

A simple Diophantine Equation, Find gcd of the coeficients and you get the answer !

The greatest common divisor of a and b is the smallest positive linear combination of a and b .

It is a known fact of the Euclidean Algorithm that the smallest positive value of $N$ such that there exists (not necessarily positive) integers $x$ and $y$ satisfying $ax + by = N$ is $\text {gcd}(a,b)$ .

Thus, the answer is $\text{gcd}(2013,3102) = \boxed{33}.$