Michael's double factorial

$\displaystyle\frac{513!}{513!!}\;=512\cdot510\cdot508\dotsc\dotsc4\cdot2\;=2^{256}\cdot(256\cdot255\dotsc\dotsc2\cdot1)$

We claim that the ratio can be written as $\displaystyle2^{a}$ $\cdot$ $\displaystyle\;b!$ in exactly two ways viz. $\displaystyle(2^{256}\times256)$ $\cdot$ $\displaystyle255!\;\;$ & $\;\;\;\displaystyle2^{256}$ $\cdot$ $\displaystyle256!$ out of which the first one gives minimum value of $b$ .

Let's say we want $a$ to be further maximum without considering the said representation. For that we would find the power of $2$ in $256!$ which comes out to be $28$ . Then we would be left with other prime factors which wouldn't form a factorial of an integer as their respective powers won't be the same.

So we conclude $\displaystyle(2^{256}\times256)$ $\cdot$ $\displaystyle255!\;\;$ is the required representation which gives $a\;=256+8\;=\boxed{264}$ .

Simplifying the fraction $\frac{513!}{513!!}$ , we will get the product of the even numbers starting from 2 to 512, which is equivalent to $512!!$ . Then, we will do some factorization by "take 2" in each term. How? We know

$512!!=512(510)(508)...(4)(2)=(2\times256)(2\times255)(2\times254)...(2\times2)(2\times1)$

Group all the 2's we had extract out earlier (except the 2 in the penultimate bracket), we arrive

$512!!=2^{256}\times 256(255)(254)...(2)(1)=2^{256}\times256!$

But $b$ has to be as small as possible and since we know $256=2^8$ , so

$512!!=2^{256}\times256\times255!=2^{256}\times2^8\times255!=2^{264}\times255!$

So, we get our desired answer which is 264.

The ratio obtained would be 2x4x6x8.......x510x512

This can be broken into 2x1x2x2x2x3x2x4........x2x255x2x256 = 2^256(256!)

256= 2^8

255 cannot be divided wholly by 2 Hence smallest value for b = 255 And consequently a = 256 + 8= 264

The given ratio $\frac{513!}{513!!}$ reduces to $2*4*6*8...512$

Now $2*4*6*8...512 = 2^{256}*256!$

And $256!=255!*2^8$

Therefore the ratio reduces to $2^{264}*255!$

We have

$\dfrac{513!}{513!!} = \dfrac{1 \cdot 2 \cdot 3 \ldots 511 \cdot 512 \cdot 513}{1 \cdot 3 \ldots 511 \cdot 513}= 2 \cdot 4 \cdot 6 \ldots 510 \cdot 512.$

We can factor $2$ from each term to obtain

$\begin{aligned} \dfrac{513!}{513!!} & = (2\cdot1) \cdot (2\cdot2) \cdot (2\cdot3) \ldots (2\cdot255) \cdot (2\cdot256) = 2^{256} \cdot (1\cdot 2\cdot3 \ldots 255\cdot256) \\ &= 2^{256} \cdot 256! = 2^{264} \cdot 255! \end{aligned}$

Note that we cannot reduce $b$ further, as 255 is not a power of 2. Hence, the corresponding value of $a$ is $\boxed{264}$ .

creary 513!/513!!=2^256*256! so the answer is 264.

$\frac{513!}{513!!}$ can be rewritten as $512!!$ since, when you expand the factorials, every other factor in the numerator cancels out with a factor in the denominator.

$512!!$ has 256 even factors. It can be rewritten as $2^{256}\times 256!$ .

The first factor of $256!$ , $256$ , is the same as $2^{8}$ . $256!$ can be rewritten as $2^{8}\times 255!$ , making the entire expression $2^{256}\times 2^{8}\times 255!$ , or $2^{264}\times 255!$ .

The exponent, $a$ , is 264 .