number of zeros

plz solve it

If n! and (4n)! end with 25 zeros and 106 zeros respectively, then which of the following is a factor of n?
a. 25
b. 26
c. 27
d. 28

‹ Sum of the digits. tricky problem ›

Submitted by Incognito on Mon, 2008-04-28 01:07.

commented on:  Mon, 2008-04-28 09:03

jigar_er_civil's picture

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ans

If n! and (4n)! end with 25 zeros and 106 zeros respectively, then which of the following is a factor of n?
a. 25
b. 26
c. 27
d. 28

if n=108 then 108! have 25 zero

and (108*4)! have 106 zero

and 108 is a factor of 27

Little Star

commented on:  Fri, 2008-05-16 05:50

Incognito's picture

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how did u arrive at 108!

how did u arrive at 108! ?
why not 105! ?

commented on:  Fri, 2008-05-16 15:10

shivanibusiness's picture

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yup, even i hv d same


yup, even i hv d same ques...

"To get the BEST, be the BEST"

commented on:  Fri, 2008-05-16 15:53

CAT@IIMs's picture

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@Shivani & Incognito

Dt no. shud satisfy two conditions...
It is evidend dt n shud be wdin 105 to 109..
n=108 only satisfies d second condn. Bcz (4*108)!=432! has 106 zeros at d end bt (4*105)!=420! has 105 zeros at d end....
So only 108! is d no....
As 108=27*4
So option (c) is correct...
Hp its clear....

Ny comments is highly accepted....

Giv ur best to d world.
Nd d best will cm back to u...

Regards,
Dipanjan.....

commented on:  Fri, 2008-05-16 17:31

shivanibusiness's picture

shivanibusiness points: 537
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I dono dipanjan...bt I am


I dono dipanjan...bt I am still not abl to get dt hw it cums b/w..105 to 109..
Sry 2 bothr u bt cn u plz elaborate d whole solution in detail..

"To get the BEST, be the BEST"

commented on:  Sat, 2008-05-17 00:26

CAT@IIMs's picture

CAT@IIMs points: 259
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Lets try, Step 1:Look n!

Lets try,

Step 1:
Look n! shud have 25 zeros in d end...
Here u have to try a bit to know the range of n...
If n=100(takin randomly keepin in view 26 is d no.of zeros]
100! has [100/5] + [100/5^2]=24 ,so n is greater than 100.
Nw as we have to increase d result by one(24+1=25) so, atleast n=105
105! has [105/5] + [105/5^2] = 25 zeros in d end
but d result wud increase by 1 if n=110
110! has [110/5]+[110/25]=26 zeros in d end
But 109! has 25 zeros in d end...
So d range of n is in between 105 to 109....

Step 2:
As d next condn. is (4n)! has 106 zeros in d end...For getting d actual value just vary d value of n wdin d range[See previous post].You will get n=108

So 108 is d no.

If still dr is ny problem dn dont hesitate..Just ask i will try my best to clarify ds..

Giv ur best to d world.
Nd d best will cm back to u.....

Regards,
Dipanjan......