Home » Ask & Discuss » Mathematics. » Algebra « Back to Discussion



Algebra

Ankit 's Avatar
Blazing goIITian

Joined: 17 Oct 2007
Post: 1659
12 Mar 2009 11:06:36 IST
0 People liked this
13
523 View Post
Zeroes
None

 How to find no. of zeroes in a factorial (not the trailing ones) ?

Example  13!


Share this article on:

Comments (13)


Blazing goIITian

Joined: 7 May 2007
Posts: 1724
12 Mar 2009 11:12:56 IST
0 people liked this

 

IN EXAMPLE 13!

WE KNOW 13!=1*2*3*4*.......*13

WE OBSERVE IN BWTWEEN THERE IS 10 .       HERE ZEROS=1

THEN WE OBSERVE THERE IS 5 WHICH WHEN MULTIPLIED BY AN EVEN NO. WILL GIVE A ZERO, MAY IT BE 2,4,6,8, OR 12.WE DO NOT CONSIDER 10 HERE BCOZ 10 HAS ALREADY BEEN USED.

SO IN 13! NO. OF ZEROS IS 2.

Ankit 's Avatar

Blazing goIITian

Joined: 17 Oct 2007
Posts: 1659
12 Mar 2009 11:22:46 IST
0 people liked this

 And find no of trailing zeroes in  101!


Blazing goIITian

Joined: 7 May 2007
Posts: 1724
12 Mar 2009 11:32:01 IST
0 people liked this

 

IN CASE OF 101!,

1)FIRST CONSIDER ALL MULTIPLES OF 10 EXCEPT 50. i.e. 10,20,30,40,60,70,80,90,100.     HERE ZEROES =10.

2)NOW CONSIDER ALL MULTIPLES OF 5 EXCEPT 10'S ,25,50,75.(i.e. 5,15,35,45,55,65,85,95)  .AND MULTIPLY THEM WITH EVEN NO. i.e. 2 .    HERE ZEROES= 8.

3) NOW  CONSIDER 25,50,75.    AND MULTIPLY  THEM WITH SOME EVEN NOS.      HERE ZEROES =6

SO TOTAL TRAILING ZEROES =10+8+6=24


Blazing goIITian

Joined: 7 May 2007
Posts: 1724
12 Mar 2009 11:48:01 IST
0 people liked this

IN CASE OF 101!,

1)FIRST CONSIDER ALL MULTIPLES OF 10 EXCEPT 50. i.e. 10,20,30,40,60,70,80,90,100.     HERE ZEROES =10.

2)NOW CONSIDER ALL MULTIPLES OF 5 EXCEPT 10'S ,25,50,75.(i.e. 5,15,35,45,55,65,85,95)  .AND MULTIPLY THEM WITH EVEN NO. i.e. 2 .    HERE ZEROES= 8.

3) NOW  CONSIDER 25,50,75.    AND MULTIPLY  THEM WITH SOME EVEN NOS.      HERE ZEROES =6

SO TOTAL TRAILING ZEROES =10+8+6=24

Anant Kumar's Avatar

Forum Expert
Joined: 10 Jul 2008
Posts: 598
12 Mar 2009 12:33:36 IST
1 people liked this

13! = 6227020800

and 101!=

9425947759838359420851623124482936749562312794702543768327889353416977599316221476503087861591808346911623490003549599583369706302603264000000000000000000000000

In both cases, Kapil's answer didn't match.


Blazing goIITian

Joined: 7 May 2007
Posts: 1724
12 Mar 2009 12:42:36 IST
0 people liked this

ANANT KUMAR SIR,

HOW ARE U SAYING THAT ANSWER IS NOT MATCHING.

IN BOTH THE CASES ANSWER IS SAME AS ACCD. TO THE SOL. GIVEN BY ME.

IN 13!, ZEROES IN THJE END =2

IN 101! ZEROES IN THE END ARE 24.

CHECK FOR YOURSELF IN THE ANSWER GIVEN BY U.

arashpreetsingh ..'s Avatar

New kid on the Block

Joined: 30 Jan 2008
Posts: 22
12 Mar 2009 12:53:17 IST
2 people liked this

to find no of zeroes simply find exponent of 5 in 101! which is[101/5] + [101/25] + [101/125]= 20+4+0=24...cheers..!! rate of if useful!
Anant Kumar's Avatar

Forum Expert
Joined: 10 Jul 2008
Posts: 598
12 Mar 2009 13:01:07 IST
1 people liked this

@kapil: the problem says the number of zeroes, NOT ONLY THE TRAILING ZEROES.

If you want to find the number of trailing zeroes in p! (where p is a prime), it is sufficient to find the exponent of 5 in p!.


Blazing goIITian

Joined: 7 May 2007
Posts: 1724
12 Mar 2009 13:21:43 IST
0 people liked this

ANANT KUMAR SIR,

 I ACCEPT MY MISTAKE REGARDING QUE. 13!,

BUT FALLING UP ASKED FOR TRAILING ZEROS IN 101!

I THINK FOR TRAILING ZEROS, METHOD IS RIGHT.

IF NOT, PLS EXPLAIN HOT TO SOLVE.

Soumik's Avatar

Blazing goIITian

Joined: 31 Jul 2008
Posts: 1266
12 Mar 2009 14:03:55 IST
0 people liked this

Dude what Anant sir said is 100% correct, finding exp. of 5 in p! does.....the trick.

Ankit 's Avatar

Blazing goIITian

Joined: 17 Oct 2007
Posts: 1659
12 Mar 2009 16:00:21 IST
0 people liked this

 Ok got the method for finding no. of trailing zeroes but how to find total no. of zeroes .

eg 13! has four zeroes.

sir, if in p! if we p is not a prime number then.

Sir how did u calculated such a long value of 101! 

pramod's Avatar

Blazing goIITian

Joined: 24 Jan 2008
Posts: 479
12 Mar 2009 18:35:51 IST
1 people liked this

jus. calculate the exponent of 5 in 101!...

ie [101/5] +[101/25]=20+4=24

now the exp of 2 in 101!..

ie [101/2]+[101/4]....+[101/64]

= 50+25+...sumthing > 24...

so we have 1024 as the power of 10...so 24 zeros..

Rohit's Avatar

Blazing goIITian

Joined: 13 Jul 2008
Posts: 382
29 Apr 2009 23:41:33 IST
0 people liked this

Then,how can we find the actual no. of zeroes in the expansion of the factorial,mathematically????



Quick Reply


Reply

Some HTML allowed.
Keep your comments above the belt or risk having them deleted.
Signup for a avatar to have your pictures show up by your comment
If Members see a thread that violates the Posting Rules, bring it to the attention of the Moderator Team
Free Sign Up!

Preparing for IIT-JEE ?

Arihant Revision Package for IIT JEE - Books, Practice Tests + Rank Predictor


@ INR 1,995/-

For Quick Info

Name

Mobile No.

Find Posts by Topics

Physics.

Topics

Mathematics.

Chemistry.

Biology

Parents

Board

Fun Zone

Sponsored Ads