Mongolian National Mathematical Olympiad

Answer: $a=2^{10}$ and $b=2^{11}$. If $b=b_{11}$ then $a_{11}=0$, which is impossible. Thus $b\ge 2b_{11}$. Let $a=a_{10}\cdot n$. Since $a>a_{10}$, we have $n\ge 2$ and $$b_{10}=a-a_{10}=a_{10}\cdot (x-1)\ge a_{10}.$$ Hence $2b_{10}\ge b_{10}+a_{10}=a$. It follows that $2b_{11}>2b_{10}\ge a\ge a_{11}$ and thus $3b_{11}>b_{11}+a_{11}=b$. This implies $b=2b_{11}$ and $a_{11}=b_{11}$. Then we have $b=2a_{11}>a>a_{10}$. This means $a=a_{11}$. Thus $a$ has 11 divisors and since 11 is a prime number, $a=p^{10}$ for a prime $p$. Then $b=2\cdot p^{10}$ and it is easy to conclude that $p=2$. Thus $a=2^{10}$, $b=2^{11}$ is the solution.