HongKong 2022-23 IMO Selection Tests

Answer: 607 Let the digits in the box form the number $\overline{abc}$. Since $22!$ is divisible by $7\times 11\times 13=1001$ and $1000\equiv -1(\mathrm{mod}1001)$, we have $$\begin{array}{rl}0& \equiv \overline{1124000727777abc680000}\\ & =1(1000{)}^7+124(1000{)}^6+727(1000{)}^4+777(1000{)}^3+\overline{abc}(1000{)}^2+680(1000)\\ & \equiv -1+124+727-777+\overline{abc}-680\\ & =\overline{abc}-607(\mathrm{mod}1001).\end{array}$$ It follows that the answer is 607.