Here is the equation $$ 6a+9b+20c=16 $$ To solve this, i follow the below steps : $\gcd(6,9)(2a+3b)+20c = 16$ let, $w = 2a+3b$ So, $3w+20c =16$ then, specific solution of $w = 112+20n$, $c = -16-3n$ ; where $n \in \mathbb{Z}$. After that, $w = 2a+3b$ So, $2a+3b = 112+20n$, I need to solve this […]

Find the number of positive integers solutions of the equation $3x+2y=37$ where $x>0,y>0,\ \ x,y\in \mathbb{Z}$ . By trial and error I found $$\begin{array}{|c|c|} \hline x & y \\ \hline 11 & 2 \\ \hline 9 & 5 \\ \hline 7 & 8 \\ \hline 5 & 11 \\ \hline 3 & 14 \\ \hline […]

Suppose $n$ be a given positive integer. Then the Diophantine equation $x=n$ has only $1$ solution. Just by inspection, I found that the Diophantine equation $x+2y=n$ has $\left\lfloor \dfrac{n}{2}+1\right\rfloor$ non-negative solutions for $(x,y).$ Also, according to this post the Diophantine equation $x+2y+3z=n$ has $\left\lfloor \dfrac{n^2}{12}+\dfrac{n}{2}+1 \right\rfloor$ non-negative solutions for $(x,y).$ Is there any closed form […]

Find all integer solutions to $7595x + 1023y=124$ Using the Euclidean algorithm I have found the $\gcd(7595,1023)=31$ and found the Bezout identity $31=52\cdot1023-7\cdot7595$ but I’m not really sure how to go about finding all solutions to that equation. I believe you can divide the equation through by the $\gcd$ – which gives $245x+33y=4$ – but […]

e.g. given two numbers $5$ and $6$ the maximum number is $19$, as after $19$ each number can be formed using equation $5n+6m$ by putting different (non-negative) values for $n$ and $m$. Such number is not possible for $4$ & $6$. I am not sure how to go-forward. The only thing I have till now […]

I want to find a set of integer solutions of Diophantine equation: $ax + by = c$, and apparently $\gcd(a,b)|c$. Then by what formula can I use to find $x$ and $y$ ? I tried to play around with it: $x = (c – by)/a$, hence $a|(c – by)$. $a$, $c$ and $b$ are known. […]

I need help with this equation: $$100x – 23y = -19.$$ When I plug this into Wolfram|Alpha, one of the integer solutions is $x = 23n + 12$ where $n$ is a subset of all the integers, but I can’t seem to figure out how they got to that answer.

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