Finding the missing digits of $23!$

Question

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.

Answer 1

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 $11$

Answer 2

Hint: $23!$ is a multiple of $99$