number of solution to the given equation. a,b,c, are all non-negative integers such that 
a + b + c=100

and 
    1000a + 300b + 50c = 10000
How many such triplets are possible? 
i have tried to reduce the equations to a relation between a b and c.
    18a + 4b = c
but this would satisfy many non negative integers. I think i am wrong somewhere. 
 A: Here's a start: It would probably help to start by noticing that the second equation is the same as
$$20 a + 6 b + c = 200$$
Using the first equation, one can write this as
$$19 a + 5b = 100$$
Now this gives some immediate bounds that will allow you to study cases: Namely, the most useful one is that $0 \le a \le 5$.
A: Solving for $a,b$ in terms of we get $$a=\frac{5c-400}{14},b=\frac{1800-19c}{14}$$
As $a$ is an integer, $5c\equiv400\pmod{14}\iff c\equiv80\equiv10$ which also makes $\displaystyle1800\equiv19c\pmod{14}$ to keep $b$ an integer
So, $\displaystyle c=14d+10$ where integer $d\ge0$
$\displaystyle\implies a=5(d-5),b=115-19d$
$\displaystyle\implies d-5=a\ge0\iff d\ge5$ and  $\displaystyle 19d-115=-b\le0\iff d\le\frac{115}{19}$
As $d$ is an integer, $d\le6$
A: You begain with these two equations:
$$
\begin{align}
a + b + c &= 100\\
1000 a + 300 b +  50 c &= 10\ 000
\end{align}
$$
First divide your second equation by 50, so that now we have:
$$
\begin{align}
a + b + c &= 100\\
20 a + 6 b +  c &= 200
\end{align}
$$
Subtracting these gives:
$$
19 a + 5b = 100
$$
Since b is non-negative, we have that
$$
\begin{align}
19 a &\leq 100\\
a &\leq 100/19 \text{ which is more than 5}\\
a &\leq 5 \text{ since it is an integer}
\end{align}
$$
But from the above equation:
$$
19 a = 100 - 5b = 5(20-b)
$$
So a must be a multiple of 5.
Hence $a=0$ or $a=5$, since it is less than or equal to 5.
If $a=0$, we get $5b = 100$, so $b = 20$. Subbing this into your first equation gives so $c = 80$.
If $a=5$ we get $95 + 5b = 100$, so $b=1$. Subbing this into your first equation gives $c = 94$.
So there are two solutions: $a = 0$, $b = 20$, $c = 80$ and $a = 5$, $b = 1$, $c = 94$.
