Should probably post this in the maths forum...
In respect to what?
Is this anything to do with binary numbers?
the method is
you need both 5 and 2 to make 10 (or 1 zero)
so start dividing by 365/5 you get 71 as the divisor ignore the remainder then divide 71/5 you get 14 ignore the remainder then 14/5 then 2/5
so now you add 71+14+2 = 87
you can check for 2 as well this number will be larger than 87, so we always choose the smaller of the two values as you can only make zero with equal 5 and equal 2
you need both 5 and 2 to make 10 (or 1 zero)
so start dividing by 365/5 you get 71 as the divisor ignore the remainder then divide 71/5 you get 14 ignore the remainder then 14/5 then 2/5
so now you add 71+14+2 = 87
you can check for 2 as well this number will be larger than 87, so we always choose the smaller of the two values as you can only make zero with equal 5 and equal 2
Original post by joy12
the method is
you need both 5 and 2 to make 10 (or 1 zero)
so start dividing by 365/5 you get 71 as the divisor ignore the remainder then divide 71/5 you get 14 ignore the remainder then 14/5 then 2/5
so now you add 71+14+2 = 87
you can check for 2 as well this number will be larger than 87, so we always choose the smaller of the two values as you can only make zero with equal 5 and equal 2
you need both 5 and 2 to make 10 (or 1 zero)
so start dividing by 365/5 you get 71 as the divisor ignore the remainder then divide 71/5 you get 14 ignore the remainder then 14/5 then 2/5
so now you add 71+14+2 = 87
you can check for 2 as well this number will be larger than 87, so we always choose the smaller of the two values as you can only make zero with equal 5 and equal 2
This method does give you the right answer, given that you can divide.
But would you like to give an explanation for the method?
This seems much like just a 'learnt' method for finding the number of zeros of a factorial.
Original post by seanahnuk
This method does give you the right answer.
Strange, I get 166 zeroes, of which 89 are trailing:
365!=
25104128675558732292929443748812027705165520269876079766872595193901106138220937419666018009000254169376172314360982328660708071123369979853445367910653872383599704355532740937678091491429440864316046925074510134847025546014098005907965541041195496105311886173373435145517193282760847755882291690213539123479186274701519396808504940722607033001246328398800550487427999876690416973437861078185344667966871511049653888130136836199010529180056125844549488648617682915826347564148990984138067809999604687488146734837340699359838791124995957584538873616661533093253551256845056046388738129702951381151861413688922986510005440943943014699244112555755279140760492764253740250410391056421979003289600000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
(edited 12 years ago)
Original post by ghostwalker
Strange, I get 166 zeroes, of which 89 are trailing:
365!=
25104128675558732292929443748812027705165520269876079766872595193901106138220937419666018009000254169376172314360982328660708071123369979853445367910653872383599704355532740937678091491429440864316046925074510134847025546014098005907965541041195496105311886173373435145517193282760847755882291690213539123479186274701519396808504940722607033001246328398800550487427999876690416973437861078185344667966871511049653888130136836199010529180056125844549488648617682915826347564148990984138067809999604687488146734837340699359838791124995957584538873616661533093253551256845056046388738129702951381151861413688922986510005440943943014699244112555755279140760492764253740250410391056421979003289600000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
365!=
25104128675558732292929443748812027705165520269876079766872595193901106138220937419666018009000254169376172314360982328660708071123369979853445367910653872383599704355532740937678091491429440864316046925074510134847025546014098005907965541041195496105311886173373435145517193282760847755882291690213539123479186274701519396808504940722607033001246328398800550487427999876690416973437861078185344667966871511049653888130136836199010529180056125844549488648617682915826347564148990984138067809999604687488146734837340699359838791124995957584538873616661533093253551256845056046388738129702951381151861413688922986510005440943943014699244112555755279140760492764253740250410391056421979003289600000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
Ok mr smarty-pants.
EDIT: They should change the word negger to nerdier solely for the benefit of the maths forum.
EDIT2: Oh, and I like being negged, it makes me more determined.
(edited 12 years ago)
Each trailing zero corresponds to a multiple of 10, so if you want to calculate the number of trailing zeros in a number whose prime factorisation is then there will be zeros.
In a factorial, there will always be more 2s than 5s in the prime factorisation (why?) so you only need to worry about 5s. You get one factor of 5 for each multiple of 5 less than or equal to 365, and an additional 5 for each multiple of 25, and an additional factor of 5 for each multiple of 125. You just need to add all these up and this gives you the number of zeros.
In a factorial, there will always be more 2s than 5s in the prime factorisation (why?) so you only need to worry about 5s. You get one factor of 5 for each multiple of 5 less than or equal to 365, and an additional 5 for each multiple of 25, and an additional factor of 5 for each multiple of 125. You just need to add all these up and this gives you the number of zeros.
Original post by DFranklin
I don't think the issue is method, more that the only person who posted a solution made an arithmetic error.
I wasn't responding to those, I was responding to the OP (who presumably hasn't really been helped by the preceding posts).
Original post by seanahnuk
This method does give you the right answer, given that you can divide.
But would you like to give an explanation for the method?
This seems much like just a 'learnt' method for finding the number of zeros of a factorial.
But would you like to give an explanation for the method?
This seems much like just a 'learnt' method for finding the number of zeros of a factorial.
I explained with the first sentence, I mentioned clearly that you need both 5 and 2 to make a 10.
If there are 100 2's and 400 5's then you can only make 200 10's.
and if you have 20 5's and 100 2's then you can make 20 10's
so when we divide a number successively by 5 we get to know how many 5's there are in that number and same is the case with 2
Yes, I learnt the method, but I understood it as well. Majority of numbers will have more 2's than 5's so we go with 5's.
go on neg me more, don't care anymore! _I_
(edited 12 years ago)
Original post by joy12
You also still seem unaware that 365/5 is not 71.
[Not that either is a big deal, but if you're going to be upset about the comments, you might as well be upset for the right reasons].
Original post by DFranklin
I think the "understanding" question is about why repeated division by 5 (and summing the quotients) gives the number of 5's dividing 365!
You also still seem unaware that 365/5 is not 71.
[Not that either is a big deal, but if you're going to be upset about the comments, you might as well be upset for the right reasons].
You also still seem unaware that 365/5 is not 71.
[Not that either is a big deal, but if you're going to be upset about the comments, you might as well be upset for the right reasons].
Thanks for pointing it out
My apologies, I made a mistake it should be 73 and not 71.
89 zero.... in 365! ...
Way to find out trailing zero
365/5= 73
365/25= 14{approx)
365/125= 2(approx)
73+14+2= 89
So there are 89 zero trailing with 365!
Way to find out trailing zero
365/5= 73
365/25= 14{approx)
365/125= 2(approx)
73+14+2= 89
So there are 89 zero trailing with 365!
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