$100! $ in terms of $2^m Z$

Question

Question

I have encountered an question. If $$ 100! = 2^m Z $$ Where $Z\notin2\mathbb Z$ is an integer, find $m$ where $m \in \mathbb{ Z^+} $

My Attempt As $$100! = 2^{50} 50!$$ $[ 1×3×5×6. . . × 99]$ $$50! = 2^{25} 25!$$ $[1×3×5. . . ×25]$

Similarly the successive terms can be written.

$ 100! = 2^{50} {2^{25}}$ (ODD term)

$100! = 2^{75} 24!$

So $$100! = 2^{97}$$ So $m = 97$ Is my approch correct ? Or it will need improvement.

Suggestions are highly appreciated.

$^*:\mathbb{ Z}^+$

Answer 1

If you need to find the exponent of a prime number $p$ in $N!$, you have to look how many times it appears (it will appear every $p$ numbers, and every $p^2$ numbers it will appear twice, and every $p^3$ numbers it will appear three times, etc).

So you are looking for the number : $$\Bigl\lfloor\dfrac{100}{2} \Bigr\rfloor+\Bigl\lfloor\dfrac{100}{2^2} \Bigr\rfloor+\Bigl\lfloor\dfrac{100}{2^3}\Bigr\rfloor+\cdots$$

which also is Legendre's Formula

Answer 2

An easier way to do this:

$$\lfloor\dfrac{100}{2}\rfloor + \lfloor\dfrac{100}{2^2}\rfloor + \lfloor\dfrac{100}{2^3}\rfloor +...$$ $$=50+25+12+6+3+1$$ $$=97$$

Answer 3

You are absolutely correct!! But lets do some smart work (instead of hard). Use Legendre's formula which states that maximum power of prime $p$, that divides $n!$ is $$\lfloor\dfrac{n}{p}\rfloor + \lfloor\dfrac{n}{p^2}\rfloor + \lfloor\dfrac{n}{p^3}\rfloor +...$$which obviously converges for sufficiently large value of $p^i$.