Romanian Mathematical Olympiad

Assume, without loss of generality, that $x < y < z < t$. Then $2^x + 2^y + 2^z + 2^t = 305 \cdot 2^a$, whence
$$2^x \cdot (1 + 2^{y-x} + 2^{z-x} + 2^{t-x}) = 305 \cdot 2^a \tag{1}$$
Since $x < y < z < t$, the numbers $y-x$, $z-x$ and $t-x$ are positive integers, hence $2^{y-x}$, $2^{z-x}$, $2^{t-x}$ are even, and $1+2^{y-x}+2^{z-x}+2^{t-x}$ is odd.
For parity reasons, (1) yields $2^x = 2^a$ and $1+2^{y-x}+2^{z-x}+2^{t-x} = 305$, hence $x=a$ and $2^{y-x}+2^{z-x}+2^{t-x} = 304$.
This gives
$$2^{y-x} \cdot (1 + 2^{z-y} + 2^{t-y}) = 2^4 \cdot 19,$$
whence $y-x = 4$, so $y = a+4$.
Moreover, $1+2^{z-y}+2^{t-y} = 19$, so $2^{z-y}(1+2^{t-z}) = 2 \cdot 9$, therefore $z-y = 1$ and $t-z = 3$, that is $z = y+1 = a+5$ and $t = z+3 = a+8$.
The above yield
$$a+x+y+z+t = a+a+(a+4)+(a+5)+(a+8) = 5a+17 = 5 \cdot (a+3)+2,$$
so the remainder of the division of the sum $a+x+y+z+t$ by $5$ is $2$.