jmc

Note that we can factor $n!+(n+1)!$ as $n!\cdot [1+(n+1)]=n!\cdot (n+2)$. Thus, we have $$\begin{array}{rl}1!+2!& =1!\cdot 3\\ 2!+3!& =2!\cdot 4\\ 3!+4!& =3!\cdot 5\\ 4!+5!& =4!\cdot 6\\ 5!+6!& =5!\cdot 7\\ 6!+7!& =6!\cdot 8\\ 7!+8!& =7!\cdot 9\\ 8!+9!& =8!\cdot 10\end{array}$$The last two numbers are $9\cdot 7!$ and $(8\cdot 10)\cdot 7!$, so their least common multiple is equal to $\mathrm{lcm}[9,8\cdot 10]\cdot 7!$. Since $9$ and $8\cdot 10$ are relatively prime, we have $\mathrm{lcm}[9,8\cdot 10]=9\cdot 8\cdot 10$, and so $$\mathrm{lcm}[7!+8!,8!+9!]=9\cdot 8\cdot 10\cdot 7!=10!.$$Finally, we note that all the other numbers in our list ($1!+2!,2!+3!,\dots ,6!+7!$) are clearly divisors of $10!$. So, the least common multiple of all the numbers in our list is $10!$. Writing this in the form specified in the problem, we get $1\cdot 10!$, so $a=1$ and $b=10$ and their sum is $\boxed{11}$.