factorials

**rubytuesday** · November 15th 2011, 05:10 PM

Solve for N:

7!8!9!/7!+8!+9!=8!N!/9

**skeeter** · November 15th 2011, 05:35 PM

Originally Posted by rubytuesday

Solve for N:

7!8!9!/7!+8!+9!=8!N!/9

noting the lack of parentheses in your equation, do you really mean this ...

$\frac{7! \cdot 8! \cdot 9!}{7! + 8! + 9!} = \frac{8! \cdot n!}{9}$

???

**Soroban** · November 15th 2011, 06:03 PM

Hello, rubytuesday!

$\text{Solve for }N:\;\;\frac{7!\,8!\,9!}{7!+8!+9!} \:=\:\frac{8!\,N!}{9}$

We have: . $\frac{8!\cdot N!}{9} \;=\;\frac{7!\cdot8!\cdot9!}{7!+8!+9!}$

. . Note that: . $7! + 8! + 9! \;=\;7!(1 + 8 + 8\!\cdot\!9) \;=\;7!(81)$

So we have: . $\frac{8!\cdot N!}{9} \;=\;\frac{\rlap{//}7!\cdot 8!\cdot 9!}{\rlap{//}7!\cdot 9^2} \;=\;\frac{8!\cdot9!}{9\cdot 9} \;=\; \frac{8!\cdot(8!\cdot\rlap{/}9)}{9\cdot \rlap{/}9}$

. . . . . . . . . . $\frac{8!\cdot N!}{9} \;=\;\frac{8!\cdot 8!}{9}$

Multiply by $\frac{9}{8!}\!:\;\;N! \;=\;8! \quad\Rightarrow\quad \boxed{N \:=\:8}$

**rubytuesday** · November 16th 2011, 02:42 AM

Thank You!

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