2
$\begingroup$

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.

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

$\endgroup$
  • 4
    $\begingroup$ $(2n)!\ne2^n\cdot n!$. $\endgroup$ – J.G. Jul 23 '19 at 14:52
  • $\begingroup$ $m\in\mathbb{I}$ ? are you sure? $\endgroup$ – Luis Felipe Jul 23 '19 at 14:53
  • $\begingroup$ @LuisFelipe Using $\Bbb I$ instead of $\Bbb Z$ for the integers is presumably a rarer convention based on English games instead of German ones. $\endgroup$ – J.G. Jul 23 '19 at 14:58
  • $\begingroup$ @J.G.yeah, I throught it was a mistake because in rarely notation $\mathbb{I}$ means irrational numbers or even pure imaginary numbers. $\endgroup$ – Luis Felipe Jul 23 '19 at 15:00
  • $\begingroup$ So for integer i have to use $\mathBbb Z $. Ok $\endgroup$ – Vedant Chourey Jul 23 '19 at 15:04
6
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
  • 4
    $\begingroup$ That's Legendre's formula. $\endgroup$ – Bernard Jul 23 '19 at 14:54
  • 1
    $\begingroup$ I got that, thankyou very much. $\endgroup$ – Vedant Chourey Jul 23 '19 at 14:56
  • $\begingroup$ @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 ;) $\endgroup$ – Luis Felipe Jul 25 '19 at 22:17
  • $\begingroup$ 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 $\endgroup$ – Vedant Chourey Jul 26 '19 at 5:00
  • $\begingroup$ just as $ \lfloor 50 \rfloor + \lfloor 25 \rfloor + \lfloor 12.5 \rfloor + \lfloor 6.25\rfloor + .... $. eventually the other terms will be 0 $\endgroup$ – Luis Felipe May 12 at 15:20
2
$\begingroup$

An easier way to do this:

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

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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 $p^i$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ And the $i$ is ? $\endgroup$ – Vedant Chourey Jul 23 '19 at 15:21
  • 1
    $\begingroup$ I just meant that for sufficiently large $i$, all terms after $\lfloor\frac{n}{p^i}\rfloor$ becomes $0$. $\endgroup$ – Anand Jul 23 '19 at 15:23
  • $\begingroup$ Ok got that. Thankyou $\endgroup$ – Vedant Chourey Jul 23 '19 at 15:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.