SAUDI ARABIAN IMO Booklet 2023

As $(10!{)}^3$ is divisible by $9$, by the divisibility criterion for the number $9$ we get $$9|4+7+7+8+4+7+2+5+8+3+a+b+7+2+0+0+0+0+0+0=64+a+b.$$ Further, by Wilson's theorem we obtain $(10!{)}^3\equiv (-1{)}^3\equiv -1(\mathrm{mod}11)$, and thus $$-1\equiv (7+8+7+5+3+b+2+0+0+0)-(4+7+4+2+8+a+7+0+0+0)=b-a(\mathrm{mod}11).$$ Since $0\le a,b\le 9$, we have $-9\le b-a\le 9$. The only number congruent to $-1$ modulo $11$ in this interval is the number $-1$, and therefore $b-a=-1$, that is, $b=a-1$. Further, we have $0\le a+b\le 18$, from which follows $$64\le 64+a+b\le 82.$$ The numbers from this interval divisible by $9$ are $72$ and $81$, and thus we get $64+a+b=72$ or $64+a+b=81$, that is, $a+b=8$ or $a+b=17$. Putting $b=a-1$ here, the first case reduces to $2a-1=8$, which is impossible. There remains the second case, $2a-1=17$, from which we get $a=9$ and $b=a-1=8$. $\square$