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.
 A: More direct path to $a+b+c = 12$:
Write $x=a+b+c$. In particular the last equation implies $31 x \ge 365$ so $x>11$. On the other hand, the first equation is $28x + 2b + 3c = 365$. If either $b$ or $c$ is nonzero, this means $28x < 364$ so $x < 13$. And $b=c=0$ is not possible because $365$ is not divisible by $28$.
added: to be fair I should complete the proof... $x=12$ means $2b + 3c = 29$. $c$ must be odd and less than $10$, so it can be $1, 3, 5, 7, 9$. Substitute in $2b+3c = 29$, then in $a+b+c = 12$ and keep only the solutions with non-negative $a$.
A: The sum of $a,b,c$ is greater than $11$ because $31(11) = 341 < 365.$
The sum has to be less than $14$ because $28(14) = 406 > 365$.
It also cannot be $13$ because, although $28(13) = 364 < 365$, we can only swap out a $30$ or $31$ for one of the $28$'s, and this puts us over $365$.
So, the sum is $12$.
The value of $c$ must be odd because the other two terms must be even, and the sum is odd.  Let's check each odd value of $c$ such that $0 \leq c \leq 12$.
$c=11$ doesn't work because an additional $28$ or $30$ puts us over $365$.
$c=9$ can work.  $31(9) = 279$, and $365-279 = 86$.  $28(2) + 30 = 86,$ so $(a,b,c) = (2,1,9)$ is a solution.
$c=7$ works a bit by inspection:  Seven months have $31$ days, four have $30$, and one has $28$, which total $365$ days.  So $(1,4,7)$ is also a solution.
$c=5$ works also.  $31(5)=155$, and $365-155= 210$, which is $30(7)$.  So $(0,7,5)$ is a third solution.
$c=3$ doesn't work because $365-3(31) = 272$, which is greater than $30(9)$.
$c=1$ doesn't work either, because $365-31 = 334 > 30(11)$.
So, three solutions:  $(2,1,9), (1,4,7), (0,7,5).$
A: 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.
A: I'm not sure what you meant by combining (2) and (2').  However, here is a possible continuation.
Equations (1) and (1') imply that $a+b+c<14$.  Equations (3) and (3') imply that $a+b+c>11$.  If $a+b+c=13$, then (2) yields $2a-3c=25$, or $a\geq 13$, whence $a=13$,  $b=0$, and $c=0$, which do not form a solution.  That is, $a+b+c=12$ must hold.  It is easy to see that $$(a,b,c)=(1,4,7)+t(-1,3,-2)$$ with $t\in\mathbb{Z}$ are the only integer solutions to $28a+30b+31c=365$ and $a+b+c=12$.  For nonnegative-integer solutions, $t\in\{-1,0,+1\}$ are the only possibilities, giving three triples $(a,b,c)=(2,1,9)$, $(a,b,c)=(1,4,7)$, and $(a,b,c)=(0,7,5)$.
