If $17! = 355687\underline{ab}8096000$. Then value of $(a,b)$ is

Solution 1:

To avoid unanswered questions, I will compile the tips from the comments here, along with a bit more.

Instead of using divisibility by $3$, use divisibility by $9,$ so that we can actually conclude that $\frac{57+a+b}{9}=6+\frac{3+a+b}9$ is an integer. Since $0\le a+b\le 18,$ then it follows that either $a+b=6$ or $a+b=15$.

As with your work, we can also conclude that $a-b=2$ or $b-a=9$. The latter, though, is impossible, since $b\le 9$ and $a\ge 0,$ so that $b-a=9$ if and only if $b=9$ and $a=0,$ in which case, $a+b\ne 6,15.$ Therefore, we have $a-b=2.$ Since $2a=(a+b)+(a-b)=a+b+2,$ then $2a=8$ or $2a=17,$ but the latter is clearly false, since $a$ is an integer. Thus, $a=4,$ so since $a-b=2,$ then solving the system gives us $(a,b)=(4,2).$

Your second question was previously answered here. Calvin Lin's answer, in particular, seems very accessible.