Finding a Diophantine Equation

To get integer solutions for the Diophantine Equation, the RHS must be divisible by the greatest common divisor of coefficient of the LHS.

$gcd$ $(225, 2325) = 75$ , therefore $82861 + N$ must be divisible by $75$ .

The smallest positive integer value satisfying is $\boxed{14}$ , as $82875$ is divisible by $75$ , giving equation $225x + 2325y = 82875$

Dividing by $75$ on both sides of the equation gives $3x + 31y = 1105$ , which certainly has many integer solutions, for example $(358, 1), (327, 4), (296, 7)$ and so on.

The Diophantine equation factors into 75(3x+31y) = 82861 + N. We know that since 3, 31 are coprime, there exist integers (x,y) such that 3x+31y=1.

Hence we seek the least positive N such that 82861+N is a multiple of 75. It follows that when N=14, 82861+14=82875 is divisible by 75.

We are given the equation $225x + 2325y = 82861 + N$ . A little bit of rearrangement gives us,

$x = \frac {82861+N-2325y}{3}$

Because $x$ is an integer, thus $82861+ N - 2325y$ must be a multiple of $225$ . Thus we have,

$82861+N-2325y \equiv 0 \pmod{225}$

or $N \equiv 2325y-82861 \pmod{225} \equiv 75y-61 \pmod{225}$

For integer values of $y$ , the only possible residues are $14,89$ and $164$ . Clearly, the minimum among these is $14$ , which is attained at $......-8,-5,-2,1,4,7.......$ .

Thus the minimum positive integer value of $N$ is $\boxed{14}$ .

Rewrite the equation as 75*(3x+31y)=82861+N, So 82861+N must be divided by 75, so the last digit of 82861+N must be 5 or 0, and 82861+14 can be divided, so N is 14.

Take any two numbers say $10$ and $20.$

No matter what combinations of integers you used for $x$ and $y$ , you can never get to the result $10x + 20y = 3.$

In fact you can only get those integers which are an integral multiple of $(10, 20) = 10.$

So let's return to the problem, we can only find those numbers which are integral multiple of $(225, 2325) = 75.$ And the nearest integer greater than $82861$ which is divisible by $75$ is $82875.$

So our answer is $82875 - 82861 = \boxed{14.}$

See also Bezout's Lemma.

Note that $225x+2325y=75(3x+31y)$ . As a result, since $\gcd(3,31)=1$ , the possible values taken on by this expression must be multiples of $75.$ The smallest multiple of $75$ greater than $82861$ is $82675$ , which gives $N=\boxed{14}$ .

(Note: one such pair that satisfies this is $(x,y)=(17,34)$ .)

Well, the gcm between $225$ and $2325$ is $75$ . Then we know that $75 | 82861 + N$ , i.e.,

$75 \times k = 82861 + N \Rightarrow 75 \times k - 82861 > 0 \therefore \frac{82861}{75} < k = 1105$

With this we have $75 \times 1105 = 82875 = 82861 + N \therefore N = 14$ .

$222x+2325y=82861+N$

Let us simplify the left hand side of this equation by dividing by $75$

$3x+31y=\frac{82861+N}{75}$

Now note that the coefficients of the left hand side are coprime. This means that, by changing the coefficients (including negatively), it is possible to generate any integer value. Therefore, all that is needed is to find the smallest value $N$ such that:

$\frac{82861+N}{75}\in\mathbb{Z}$

Note:

$82861\div75=1104.81\dot{3}$

Thus the smallest value of N occurs when:

$82861+N=75\times1105$

$82861+N=82875$

$\boxed{N=14}$

A linear Diophantine equation of the form ax + by = c has solutions if and only if gcd(a,b) | c. gcd(225,2325) = 75. As 5 | L.H.S. so 5 | R.H.S. This eases our calculation to find N and we get N = 14.

25 divides LHS so we have that N must be congruent to 14 module 25. The smallest multiple of 25 bigger than 82861 is 82875. So the smallest possible N is 14.