75th 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 $$2^x\cdot (1+2^{y-x}+2^{z-x}+2^{t-x})=305\cdot 2^a\text{\hspace{1em}(1)}$$ 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 $2$.