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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.

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    $\begingroup$ Another way to solve it would be just to compute the factorial ... $\endgroup$
    – hmakholm left over Monica
    Commented Aug 21, 2015 at 17:13
  • $\begingroup$ @HenningMakholm-This though came into my mind when I saw it at first but the method is not useful if it given to you when you do not posses a calculator (maybe during an exam). $\endgroup$
    – Soham
    Commented Aug 21, 2015 at 17:15
  • $\begingroup$ I don't have a calculator, but I do have a computer... perl -e 'system join "*","calc 1",2..23' $\endgroup$
    – hmakholm left over Monica
    Commented Aug 21, 2015 at 17:17
  • $\begingroup$ @HenningMakholm-What about during an exam?:-) $\endgroup$
    – Soham
    Commented Aug 21, 2015 at 17:18
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    $\begingroup$ Why wouldn't you use divisibility rules? Henry's shortcut is great, but even if that doesn't occur to you, the rules for 9 and 11 leave you with $\left\{\begin{aligned}x+y&=4\text{ or }13\\x-y&=-1\end{aligned}\right.$ which is quick to solve. $\endgroup$
    – 2'5 9'2
    Commented Aug 21, 2015 at 17:33

2 Answers 2

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

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  • $\begingroup$ How do you want to continue? $\endgroup$
    – Soham
    Commented Aug 21, 2015 at 17:18
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    $\begingroup$ This is a great hint. +$1$. $\endgroup$
    – Clayton
    Commented Aug 21, 2015 at 17:18
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    $\begingroup$ @tatan: It's easy to compute residues modulo 99 if you go to base 100... $\endgroup$
    – hmakholm left over Monica
    Commented Aug 21, 2015 at 17:21
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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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    $\begingroup$ I can solve it by using divisibility rules.But as I mentioned in the question,I am looking for a different method. $\endgroup$
    – Soham
    Commented Aug 21, 2015 at 17:17
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    $\begingroup$ If you were using divisibility rules, I'd also use the one for $11$ and the one for $9$. $\endgroup$
    – 2'5 9'2
    Commented Aug 21, 2015 at 17:18
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    $\begingroup$ considering $3$ is of couse redundant if you consider $9$ anyway ... $\endgroup$
    – Hagen von Eitzen
    Commented Aug 21, 2015 at 17:32

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