Find the largest prime number p such that when 2012! is written in base p, it has at least p trailing zeroes.

mathhdude124 May 9, 2019

#1**+1 **

Find the largest prime number p such that when 2012! is written in base p, it has at least p trailing zeroes.

If you write any number in a different base then the number of zeros will be the number of times that base can be divided into the number

For example. 20 written in base 2 will have 2 zeros at the end because 20 can be divided by 2 twice and get an integer answer.

I have deleted what I wrote after this point because it was nonsense. BUT

You should be able to use this initial fact to help determine the correct answer.

Melody May 12, 2019

#3**0 **

Yes, you are absolutely correct. Great observation!

My last sentence is completely wrong. (I willl delete it with an explanatory note)

I think you can still use the beginning of my explanation to determine the correct answer. I may do it myself when I have a little more time.

Melody May 12, 2019

#4**+2 **Find the largest prime number p such that when 2012! is written in base p, it has at least p trailing zeroes.

**Find the largest prime number p such that when 2012! is written in base p, it has at least p trailing zeroes.**

The answer is 43 because 43^2 is the biggest squared prime number that is less than 2012!

I did work this our logically but BUT I will admit that it may be difficult for me to show this logic.

I will try

Say the number was not 2012! Say it is what ever number factorial that will work if

**p is 2**

1*2*3*4 = 1*2*3*2*2 = 3*2*2*2 this will have 3 trailing zeros when written in base 2 (there is no version with only 2 trailing zeros)

so if p=2 then it would have to be factorial 4. Note that 2^2=4

**Try p=3 **

I need the smallest factorial (X!) where X! written in base 3 has at least 3 trainling zeros. So I need at least 3^3 = 27 to be a factor

1*2*3*4*5*6*7*8*9 = 1*2*3*4*5*2*3*7*8*3*3 = 3^4 * 2*4*5*2*7*8 = 9!

when 9! is written in base 3 there will be 4 trailing zeros .. It is imposible to have only 3

**Try p=5**

I need the smallest factorial (X!) where X! written in base 5 has at least 5 trainling zeros. So I need at least 5^5 to be a factor

1*2*3*4*5*6*7*8*9*10* ..... 25

= (Some integer without 5 as a factor)* 5*1* 5*2 * 5*3 * 5*4 * 5*5

= (Some integer without 5 as a factor)* 5 * 5 * 5* 5 * 5*5

= (Some integer without 5 as a factor)* 5^6

when 25! is written in base 5 there will be 6 trailing zeros .. It is imposible to have only 5

Let me try and explan the general version.

Let the number be p

Consider this

\((p^2)!=1*2*......1p.......*2p........*(p-1)p.....*(pp) \\ (p^2)!=1*2*...1*2* ...*(p-1).....*p^{(p-1)}.....*p^2\\ (p^2)!=(\text{An interger without p as a factor}).....*p^{(p-1)}.....*p^2\\ (p^2)!=(\text{An interger without p as a factor}).....*p^{(p+1)}\\\)

Back to the original question

\(\sqrt{2012}\approx 44.8\)

The closest prime number less than 44.8 is 43

**So p=43**

**-------------------------------------------------------------------------**

Lets check

\(2012! = \text{(Some integer that does not have 43 as a factor)}*43*2*43*3*43 *..........42*43*43*43\\ 2012! = \text{(Some integer that does not have 43 as a factor)}*43^{42}*43*43\\ 2012! =\text{(Some integer that does not have 43 as a factor)}*43^{44}\\ \text{When written in Base 43 2012! will have 44 trailing zeros.}\)

What about if p=47, maybe it will work too?

\( 2012! = \text{(Some integer that does not have 47 as a factor)}*47*2*47*3*47 *..........42*47\\ 2012! = \text{(Some integer that does not have 43 as a factor)}*47^{42}\\ \text{When written in Base 47 2012! will only have 42 trailing zeros.}\\ \text {so Base 47 will definitely not work} \)

Melody May 13, 2019

#5**+1 **

Hello Melody: I wrote a very short code to convert 2012! to base 43 in its entirety: Here are just the beginnings of the number and its trailing zeros. They count to 47 zeros.

n=2012!;m=43; a=n%m;printa,", ",;b=int(n/m);n=b;if(n>m, goto2, discard=0;printb,

(6, 38, 32, 10, 35............ 14, 21, 18, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0)

P.S. I also converted it to base 47 and it gives 42 zeros.

Guest May 13, 2019

edited by
Guest
May 13, 2019

#6**+1 **

**Note: To count the number of trailing zeros of N! in any base above 36, simply divide N by the powers of the base that are <= N.**

**Example: 2012! in base 43, we simply divide:2012 / 43^1 + 2012 / 43^2 =46 + 1= 47 zeros. Just take the INTEGER part.**

**2012! in base 47: 2012 / 47^1 + 2012 / 47^2 = 42 + 0 = 42 zeros. [47^2 is > than 2012, hence = 0]**

Guest May 13, 2019

#7**+2 **

Yes that sounds reasonable.

You say there are 47 trailing zeros

Our number is 2012!

43^2 = 1849

Since 1849 is smaller than 2012 so I would expect there to be more than 44 trailing zeros when 2012! is written in base 43.

I know I have told you many times before but I really like your little lines of code that you figure out and run.

It is a really helpful check mechanism.

I'd really like it if you became a member.

You could call yourself CodeDude OR Cplusplus ( I think C++ is the code you use??) LOL

OR you could chose whatever you want.

Melody May 14, 2019