Mongolian Mathematical Olympiad

Answer: $\{a,b,c,d\}=\{4,5,6,7\}$. It is clear that the above is a solution, so we prove that there are no other solutions. We may assume $a\le b\le c\le d$. Since $S=2^a+2^b+2^c+2^d=15\cdot 4a$, we have $2^a\mid 4a$, thus $2^a\le 4a$ and therefore $a\le 4$.

First, we prove that if $S\equiv 0(\mathrm{mod}15)$, then $a,b,c,d$ must have distinct remainders modulo $4$. Since $2^4\equiv 1(\mathrm{mod}15)$, we may order the remainders as $0\le p\le q\le r\le s\le 3$ and furthermore we may assume that $p=0$. Then we have $s=3$, since $2^0+3\cdot 2^2<15$. Similarly, $r=2$, since $2^0+2\cdot 2^1+2^3<15$. Finally, it is clear $q=1$.

It follows that $v_2(S)=a=2+v_2(a)$, hence $a\ne 1,2,3$. For $a=4$, we have $b\ge 5$, $c\ge 6$, $d\ge 7$, thus $S\ge 240$. Equality means there is no other solution.