SELECTION and TRAINING SESSION

Answer: $2310$. Suppose $a$ and $b$ satisfy the problem conditions. For each integer $k$ from $1$ to $10$ denote by $p_k$ the minimal prime divisor of $\gcd (a+k,b+k)$. Since the $\gcd$ of numbers divides their difference, among $10$ successive integers at most $\lfloor \frac{10}{p_k}\rfloor$ can be a multiple of $p_k$. For $p_k=2$ this number is $5$, for $p_k=3$ it is $4$, for $p_k=5$ and $p_k=7$ it is $2$ and for the others it equals $1$. If all $p_k$ are odd then the equality $10=4+2+2+1+1$ implies that there is at least five distinct primes. Then $b-a\ge 3\cdot 5\cdot 7\cdot 11\cdot 13>2310$. Consider the case when some $p_i=2$. Since $a$ and $b$ are coprime, $i$ cannot be even so $2=p_1=p_3=\cdots =p_9$. Among five primes $p_2,p_4,\dots ,p_{10}$ at most $2$ can be equal to $3$, and for each prime $p\ge 5$ at most one can be equal to $p$. Now the equality $10=5+2+1+1+1$ implies that $b-a\ge 2\cdot 3\cdot 5\cdot 7\cdot 11=2310$. On the other hand, if $a=2311$ and $b=4621$ then $p_1=p_3=\cdots =p_9=2$, $p_2=p_8=3$, $p_4=5$, $p_6=7$ and $p_{10}=11$, whence $|a-b|=2310$ is possible.