Cfrtjryk

Prove that for any integer value of $D$, the equation $27x + 14y = D$ has integer solutions for $x$ and $y$.

$$27x+14y=D(28-27)$$

$iff27(x+D)=14(2D-y)$

$dfrac{14(2D-y)}{27}=x+D$ which is an integer

$implies27|14(2D-y)implies27|(2D-y)$ as $(14,27)=1$

$dfrac{2D-y}{27}=c$(say) where $c$ is an arbitrary integer

$implies y=?$

$implies x=?$

If you can find $27x_0 + 14y_0 = 1$ then can find $27(D*x_0) + 14(D*y_0) = D$.

I'm by no means a math expert, but it seems to me that if you can solve for D=1, (x = -1, y = 2), then multiplying the entire equation by any integer, results in an integer solution for the general equation. I don't know how to put this into mathematical proof terms, but basically because there is a solution where D = 1, then multiplying the entire equation by some arbitrary integer c means that for any integer D, there is a solution, because you can multiply both x and y by the same number, and get a solution.

Maybe someone else can give a more formal explanation of what I'm trying to say.

$$27x + 14y = D$$

The first step is to find one solution to $27x + 14y=1$

An "obvious" solution is $(x,y)=(-1,2)$.

Assuming you want to have a general method for finding solutions to such problems...

Start with
begin{array}{c} 27 = 27(1) + 14(0) \ 14 = 27(0)+14(1) end{array}

The idea is to manipulate "things" so that the number on the left becomes a $1$.

For example, $13 = 27 - 14 = 27(1-0) + 14(0-1)= 27(1) + 14(^-1)$.

We end up with the list

begin{array}{rcl}
27 &= &27(1) + 14(0) \
14 &= &27(0)+14(1) \
13 &= &27(1)+14(^-1)
end{array}

Next we see that $1=14 - 13 = 27(0-1)+14(1-(^-1))=27(^-1)+14(2)$. So the list looks like

begin{array}{rcl}
27 &= &27(1) + 14(0) \
14 &= &27(0)+14(1) \
13 &= &27(1)+14(^-1) \
1 &= &27(^-1)+14(2)
end{array}

Next we find a solution to $27x + 14y=D$

Since $27(^-1)+14(2)=1$, then $27(-D)+14(2D)= D$

Finally, we solve $27x + 14y=D$

Suppose that $27x + 14y=D$ for some $x$ and $y$. Then
begin{align}
27x + 14y=D &= 27(-D)+14(2D)= D \
27(x+D) &= 14(2D-y) \
end{align}

Since $27 mid 27(x+D)$, then $27 mid 14(2D-y)$.

Since $gcd(27,14)=1$, then $27 mid 2D-y$.

Hence, for some integer, $t$

begin{align}
2D - y &= 27t \
y &= 2D-27t
end{align}

Next, we solve for $x$

begin{align}
27x + 14y &= D \
27x + 14(2D-27t) &= D \
27x + 28D - 14(27)t &= D \
27x &= 14(27)t - 27D \
x &= 14t - D
end{align}

So the general solution is

$$(x,y) = (14t-D, 2D-27t)$$

for all integers, $t$.

I see you have already some interesting answers and I will try another way to explain.

Let us make prime number factorization of numbers $27$ and $14$:

$$27 = 2^0times 3^3times 7^0\14=2^1times3^0times 7^1$$

They don't have any non-zero exponent for the same prime base. This means $27$ and $14$ are relatively prime. If they were not relative prime, they would have some common factor $K>1$ and we could write $$K(ax+by)=D$$
But since any number $D$ is not divisible by any given common factor $K>1$, we are sure to be able to hit it.

An example if we did not have relative prime numbers is $$27x+15y=D Leftrightarrow 3(9x+5y)=D$$
Which we can see that it could only be sure to fit if $D$ was divisible by $3$.