The only solution of the equation ${72_8!}/{18_2!}=4^x$ is $x=9$

Question

Problem and Definitions

If $n_a!:=n(n-a)(n-2a)(n-3a)\ldots(n-ka):n>ka$, how should I go about solving this?: $$\dfrac{72_8!}{18_2!}=4^x$$

Attempt

$$\dfrac{72(72-8)(72-16)(72-24)(72-32)(72-40)(72-48)(72-56)(72-64)}{18(18-2)(18-4)(18-6)(18-8)(18-10)(18-12)(18-14)(18-16)}=4^x$$ which is then simplified into (execution) $$\dfrac{72(64)(56)(48)(40)(32)(24)(16)(8)}{18(16)(14)(12)(10)(8)(6)(4)(2)}=4^x$$ Now we see there is a bijective function from the multiplicands; $72=18\times 4$, $56=14\times 4$, $64=16\times 4$, $56=14\times 4$, $48=14\times 4$... So we can simplify to, by dividing each top be the bottom: $$\dfrac{4(4)(4)(4)(4)(4)(4)(4)(4)}{1}=4^x$$ to $$4^24^24^24^24^1=4^x$$ to $$4^{2+2+2+2+1}=4^x$$ then to $$4^{9}=4^x.$$ From this, we can obviously see that $x=9$.

Question

Am I correct in assuming that there is a bijective function between every dividend and divisor? And, am I also correct in taking each dividend and dividing it by its respective divisor? Generally, is this 'proof' correct?

Answer 1

Your reasoning is fine. You don't need a bijective function. What you are doing is dividing the numerator and denominator by the same amount a bunch of times. As you say, this is made easier because you can divide each factor in order by the other, resulting in a quotient of $4$, then just count up the factors of $4$.

Answer 2

Note that if $a|n$, as it does in your case, then, if $b = \frac{n}{a}$,

$\begin{array}\\ n_a! &=\prod\limits_{k=0}^{b-1} (n-ka)\\ &=\prod\limits_{k=0}^{b-1} (ba-ka)\\ &=a^b\prod\limits_{k=0}^{b-1} (b-k)\\ &=a^b\prod\limits_{k=1}^{b} k\\ &=a^b b!\\ &=a^{n/a} (n/a)! \end{array} $

Therefore $\dfrac{72_8!}{18_2!} =\dfrac{8^9 9!}{2^9 9!} =\dfrac{8^9}{2^9} = 4^9 $.

Answer 3

Short solution (Ross answer as a formula) $$ \dfrac{9\cdot 8 - 8k}{9\cdot 2 - 2k}=\dfrac{8\cdot(9-k)}{2\cdot( 9 - k)}=4 \qquad (0\leq k\leq 8) $$