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

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.

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

• $(2n)!\ne2^n\cdot n!$. – J.G. Jul 23 '19 at 14:52
• $m\in\mathbb{I}$ ? are you sure? – Luis Felipe Jul 23 '19 at 14:53
• @LuisFelipe Using $\Bbb I$ instead of $\Bbb Z$ for the integers is presumably a rarer convention based on English games instead of German ones. – J.G. Jul 23 '19 at 14:58
• @J.G.yeah, I throught it was a mistake because in rarely notation $\mathbb{I}$ means irrational numbers or even pure imaginary numbers. – Luis Felipe Jul 23 '19 at 15:00
• So for integer i have to use $\mathBbb Z$. Ok – Vedant Chourey Jul 23 '19 at 15:04

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

• That's Legendre's formula. – Bernard Jul 23 '19 at 14:54
• I got that, thankyou very much. – Vedant Chourey Jul 23 '19 at 14:56
• @VedantChourey if your questions has been solved, don't forget to mark an answer as "correct answer" for helping other users with your same question to find an answer ;) – Luis Felipe Jul 25 '19 at 22:17
• How to do this operation? Actually i'm operating the site through an android application, so i don't know how to perform the above mentioned action. Please give direction – Vedant Chourey Jul 26 '19 at 5:00
• just as $\lfloor 50 \rfloor + \lfloor 25 \rfloor + \lfloor 12.5 \rfloor + \lfloor 6.25\rfloor + ....$. eventually the other terms will be 0 – Luis Felipe May 12 at 15:20

An easier way to do this:

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

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 $$p^i$$.

• And the $i$ is ? – Vedant Chourey Jul 23 '19 at 15:21
• I just meant that for sufficiently large $i$, all terms after $\lfloor\frac{n}{p^i}\rfloor$ becomes $0$. – Anand Jul 23 '19 at 15:23
• Ok got that. Thankyou – Vedant Chourey Jul 23 '19 at 15:30