The number of pairs of integers $(x,y)$ satisfying $x ≥ y ≥ -20$ and $2x + 5y = 99$ is I tried to solve this question but I was unable to think how to get the number of integer pairs satisfying these conditions. Till now I have broken down this into :-
$2x+5y = 99$
==>$y=(99-2x)/5$
$y\geq-20$
==> $(99-2x)/5 \geq20$
==> $x\leq 99.5$
$x\geq y$
==> $x \geq (99-2x)/5$
==> $x\geq 99/7$
From here I am not able to process further to think of the way that will give me the solution.
 A: $2x + 5y = 99$
$x \geq y \geq - 20$
If $x = y, 7 x = 7 y = 99$ which means for $x \leq 14$, $y \gt x$ so we will have solutions only for $x \gt 14$.
Also, $5y \geq -100 \,$ so $\,2x \leq 199 \implies x \lt 100$
Also, $2x$ is an even number so the only way for $y$ to be integer is when $2x$ ends in $4$ which means $x$ ends in either $2$ or $7$.
So, $x = 17 + 10m, 22 + 10n \lt 100$ ($m, n \geq 0) \,$are the solutions. You can easily count the solutions.
A: $2x + 5y = 99$ is very rare and a strong restriction.
If $2a + 5b=99$ is a solution to find another solution $2(a\pm d) +5(b\mp e) =99=2a+5b$ we must have $2d - 5e = 0$ and that requires that $d$ be divisible by $5$ and $e$ be divisble by $2$.
so if $2a + 5b =99 $ is one answer than any $2(a + 5k) + 5(b-2k) =99$ is also a solution and these are the only solutions.
Now lets find our first solution.
$2x +5y = 99$ so $5y = 99-2x$ and we need the RHS to be divisible by $5$.  We can do that by letting $x = 2$ and $5y = 95$ so $y = 19$.
So if $2x + 5y = 99$ then $x = 2+5k$ and $y = 19 - 2k$ for some integer $k$ and for any integer $k$ $x = 2+5k$ and $y =19-2k$ will be a solution.
So we must have $2+5k \ge 19-2k\ge -20$.  How many integers key can that be true for?
We must have $2+5k \ge 19-2k$ so $7k \ge 17$ so $k \ge 3$.  And we have $19-2k \ge -20$ so $2k \le 39$ so $k \le 19$.  So that's it.
$3\le k \le 19$ and $x = 2+5k$ and $y = 19 - 2k$ are all the solution pairs.  There are $17$ of them:
$(x,y) = (17, 13),(22,11),(27,9).....,(97, -19)$.
All other solutions will either have $y > x$ (such as $(12,15)$) of have $y < -20$ (such as $(102, -21)$.
A: y must be of the form $y=-(2k+1)5\geq-20$ which gives:
$(2k+1)5= (-5, -15) \geq -20$ $\rightarrow$ $2k+1=1, 3$
So for any $k\geq 0$ there will be an integer solution for x. in fact number of solutions is 2 when y is negative. Examples:
$2k+1=1$ $\rightarrow$ $y=-5>-20$, gives : $2x=25+99$ $\rightarrow$ $x=62$
$2k+1=3$ $\rightarrow$ $y=-15>-20$ ,gives: $2x=99+75$ $\rightarrow$ $x=87$
Now suppose x is negative then we must have:
$2x=-4, -14, -24, -34, -44, -54, -64, -74, -84, -94$
And corresponding y is:
$y=19, 17, 15, 13, 11, 9, 7, 5, 3, 1$
But this is not possible because dute to statement $x>y\geq-20$.
Now suppose x and y are both positive we have:
$x=47, 42, 37, 32, 27, 22,17, 12, 7, 2$
$y=1, 3, 5, 7, 9, 11, 13, 15, 17, 19 $
In this case only following solutions safices the condition:
$(x, y)=(47,1),(42, 3), (37, 5), (32, 7), (27, 9), (22, 11), (17, 13)$
Hence number of solutions is 9.
A: This is a Linear Diophantine equation. As the first step we need to check whether the greatest common divisor (GCD) of 2 and 5 divides 99 (it does). Once we have done that we need to find a way to express the GCD of 2 and 5 as $2x_0+5y_0 = 1$, where 1 is their GCD and $x_0$, $y_0$ are integers. $x_0$ and $y_0$ can be found using Euclidean Algorithm, in this case $x_0=-2$ and $y_0=1$. Now we multiply both the sides of the equation by 99 to get $99=2(-198)+5(99)$. Consequently, we set $x_0=-198$ and $y_0=99$. Voila! You have found the answer! Oh, not yet. We need to fix few things. We have to keep the constraints in mind. Here comes the general solution to our equation, $x=x_0+5t$ and $y=y_0-2t$, here (x,y) is our general solution. We apply the given inequalities on x and y to get the following range for t (remember that t is an integer), $43\leq t \leq 59$. So the number of possible integral values of t is $59-43+1=17$, and corresponding to each value of t we have a different solution, so there are a total of 17 solutions.
A: It's much easier if we restrict variables to be non-negative.
So we have $y=z-20, x=z+w-20, z\ge 0, w \ge 0$
$2x+y=99 \implies 2(z+w-20)+5(z-20)=99 \implies 7z+2w=239$
Then $z\equiv 1 \pmod 2 \implies z=2m+1, m \ge 0$
$2w\equiv 1 \pmod 7  \implies w\equiv 4 \pmod 7 \implies w=7n+4, n\ge 0$
Finally, $7(2m+1)+2(7n+4)=239 \implies m+n=16$
So there are 17 solutions.
