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Number System Problem
by Gaurab Pantu - Saturday, 7 June 2008, 10:48 AM
 

1)What is total number of Zero at the end of (625!)!.

2)What is the total number of digits of 1!+2!+3!+.......+98!+99!+100!

3)Last Two digits of 8^7^6^5^4^3^2.

 

 

Re: Number System Problem
by King Anand - Saturday, 7 June 2008, 04:41 PM
  3.36......
Re: Number System Problem
by kali dada - Saturday, 7 June 2008, 05:25 PM
 

But how? Please explain in detail.

 

Re: Number System Problem
by King Anand - Saturday, 7 June 2008, 08:24 PM
 

sorry i made a mistke in solvn...... not the answer...sad ...

i think ......      08.....   but pehle plzzz say rite or not...then i'llpost my soln..

Re: Number System Problem
by Sourojit Bhattacharya - Monday, 9 November 2009, 09:46 PM
  Hi,

I believe the answer should be 128

Explanation
For those number divisible by 5 must be ended with a zero when it is multiplied by an even number

number of zeroes ----> 125 ---------  (i)
For those number who are divisible by 25 will give 2 zeroes so 25 more zeroes added ----- > 25 --------- (ii)

number divisible by 125 will give us one more zero than divisible  by 25
3 added ----- (iii)
By adding I + ii +iii
So answer is 153

Correct me if I am wrong


Re: Number System Problem
by ritesh chopra - Thursday, 1 April 2010, 09:43 PM
  no in such kind of ques u just have to find no of 5 in 625!
for that jus keep on dividing by 5...
like 625/5 = 125
125/5 =25
25/5 =5
5/5 = 1
now the ans is 125+25+5+1=156
Re: Number System Problem
by rahul sharma - Thursday, 8 April 2010, 05:35 PM
  156 is alryt..but that is only for single factorial ,wat about d other one..
i think it will be.. 194...
correct me if m wrng
Re: Number System Problem
by rahul sharma - Thursday, 8 April 2010, 05:41 PM
  last two digits will be 28 following general method of cyclicity.
need to dig up for 2nd one..
Re: Number System Problem
by Harmit Choudhary - Monday, 21 March 2011, 12:14 PM
  How to proceed in the 2nd que?
Re: Number System Problem
by TG Team - Wednesday, 23 March 2011, 01:32 PM
 
Hi Harmit smile

I hope you can easily calculate the number of digits of 10100 as 100 + 1 = 101 or more mathematically it is [log10100] + 1.

Now in our case if we can find log(100!), then things can be simplified.
For our goodness, there is one good approximation to this as follows:
log(n!) ~ n(log(n) - 0.43) + 1

By putting, n = 100, we get log(100!) = 100(log100 - 0.43) + 1 = 158.
So 100! contains 159 digits approximately. smile

Kamal Lohia
Re: Number System Problem
by Rajasekaran Rajaram - Friday, 27 May 2011, 10:03 PM
  kamal,

Is this approximation applied to find the number of digits of 100! or the number of digits in the sum of 1!+2!+...100!.
Re: Number System Problem
by VINIT ANAND - Monday, 18 July 2011, 10:57 PM
  kamal sir ,we are waiting for your answer to 3rd ques....plzz help us out
Re: Number System Problem
by Ravi Kumar - Tuesday, 4 September 2012, 07:30 PM
  Q1- its simple i think
625/5=125
625/25=25
625/125=5
625/625=1
so total 125+25+5+1=156

Q2-Im not sure about this question but a sequence is being followed in this question
The sum of 1!+2!+3!+4!+5!= 3 Digit no
The sum of 6!+7!+8!+9!+10!= 7 Digit no
The sum of 11!+12!+13!+14!+15!= 13 Digit no
The sum of 16!+17!+18!+19!+20!= 20 Digit no

so its making a series so total no of digits in 1!+2!+3!.....+100! should be 258

Q3- Here also a sequece is followed:
        08       [1st]
08*8=64
64*8=512
12*8=4096     [4th]
96*8=__68   
68*8=__44
44*8=__52
52*8=__16    [8th]
16*8=__28  
28*8=__24
24*8=__92
92*8=__36    [12th]
36*8=__88  
88*8=__04
04*8=__32
32*8=__56    [16th]
56*8=__48  
48*8=__84
84*8=__72
72*8=__76    [20th]
76*8=__08
So finally we have found a cycle of 20.
Simplifying the question 8^7^6^5^4^3^2 =8^5040
5040 mod 20 = 0

SO last 2 digit's will be 76.

Correct me if method is wrong. Its a lengthy question.
Re: Number System Problem
by DeeScript .in - Sunday, 9 June 2013, 08:01 AM
  q.1...
the total number of zeros will be the total number of 10s.
 and the toal number of 10s will he total pairs of 2 and 5
we can see tha  total number of 2 in 625 is more thant that of total numbers of 5....hence we will calculate the total number of 5s only

i.e 625/5=125
125/5=25
25/5=5
5/5=1

by adding we get total number of 5s =>156
hence the number zeros will be 156.(because there will be 156 pairs of 5 and 2).

q.3

the last 2 digits i got here is 8 ..please correct me if i am wrong..
Re: Number System Problem
by Ankurr Kumar - Saturday, 10 August 2013, 12:03 AM
  156 zeros