<s>[INST] Finding the missing digits of $23!$ It is given $23!=2585201xy38884976640000$. Now it is required to find the value of $x$ and $y$. I know I could find it by using divisibility rules and solving simultaneous equations. Is there any other way to solve it (without computing it by a calculator)? This question is just out of curiosity.
 [/INST] Given $23!=2585201xy38884976640000$, Now Here $23!$ must Contain a no. $3,9$
So $\bf{R.H.S}$ must be divisible by  $3$ and $9$
If no. is Divisible by $3\;,$ Then Sum of Digit on $\bf{R.H.S}$ is divisible by $3$
So $2+5+8+5+2+0+1+x+y+3+8+8+8+4+9+7+6+6+4$ must be divisible by $3$
So $88+x+y$ is divisible by $3$
So $1+x+y$ must be divisible by $3$
So $x+y = 2,5,8,11,14,17$
Similarly  If no. is Divisible by $9\;,$ Then Sum of Digit on $\bf{R.H.S}$ is divisible by $9$
So $2+5+8+5+2+0+1+x+y+3+8+8+8+4+9+7+6+6+4$ must be divisible by $9$
So $88+x+y$ is divisible by $9$
So $7+x+y$ must be divisible by $9$
So $x+y = 2,11$
Now also Divisibility test for $11$. If no. is divisibility by $11$
Then $\displaystyle \bf{(Sum \; of odd\; position \; no)-(sum\; of \; evev \; position\; no.)}$ must be divisible by $11$
So $(48+y)-(38+x) = 10-(y-x)$ is divisible by $11$
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