If $28a + 30b + 31c = 365$, then what is the value of $a +b +c$? Question:
For 3 non negative integers $a, b, c$; if $28a + 30b + 31c = 365$ what is the value of $a +b +c$ ?
How I approached it :
I started immediately breaking it onto this form on seeing it :
$28(a +b +c) +2b +3c  = 365 .......(1)$
$30(a +b +c) -2a +c  = 365 .......(2)$
$31(a +b +c) -b -3a = 365 .......(3)$
And then I find out that 
$365 = 28*13 + 1......(1')$;
$365 = 30*12 + 5......(2')$
$365 = 31*11 + 24......(3')$
Now as we see (1) and (1') as well as (3) and (3') or even equations $2$ and $2'$ do not combine  quiet congruently, so I meet with a dead end here.
my issue :
how should I approach such problems where we are given no other equations or data? Basically I am asking what are a few ways to get a solution for this problem.

Your equation (2) should say
$$ 30(a+b+c) - 2a + c = 365. $$
Note that this implies
$$ c- 2a \equiv 5 \mod 30. $$
I now claim that, in fact, $c-2a = 5$. This would imply that $30(a+b+c)+5=365$, or that $a+b+c = 12$.

Since we know that $c-2a\equiv 5\mod 30$, it suffices to show that $-25< c-2a < 35$ to conclude that $c-2a = 5$. The upper bound is easy, since 
$$31c\le 365\implies c\le 11\implies c-2a\le 11 < 35$$
as $a$ is nonnegative. To get the lower bound, we note that
$$ 28a\le 365\implies a\le 13.$$
Furthermore, if $a=13$, then $28\times 13 + 30b+31c = 365\implies 30b+31c = 1$, which is clearly impossible for nonnegative $b$ and $c$. Thus, $a\le 12$, and hence
$$ c-2a\ge 0 - 2(12) = -24 >-25 $$
since $c$ is also nonnegative. Hence, $-25 < c-2a < 35$, and combined with the fact that $c-2a\equiv 5\mod 30$, we conclude that $c-2a = 5$, as desired.

Using the fact that $c-2a = 5$, we can also solve for the possible solutions. Substituting $c = 2a+5$ into the original equation yields
\begin{align} 28a+30b+31(2a+5) &= 365 \\
\implies 28a + 30b + 62a + 155 &= 365 \\
\implies 90a + 30b &= 210 \\
\implies 3a + b &= 7.
\end{align}
We thus see that
\begin{align} a = 0 &\implies b = 7, c = 5 \\
a = 1 &\implies b = 4, c = 7 \\
a = 2 &\implies b = 1, c = 9.
\end{align}
If $a\ge 3$, then $b<0$. Hence, we conclude that $(0,7,5)$, $(1,4,7)$, and $(2,1,9)$ are the only solutions to the problem.
