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Factorials
by Software Engineer - Wednesday, 17 June 2009, 04:53 PM
  cat 2009 cat 2010 number system factorialMost of the CAT aspirants don’t know that the hardest work they have to do for their CAT preparation is developing a healthy psychology. What is a healthy psychology? That you feel good about yourself, your family, your relationships, your academics, your job, your friends, everything. Which means that you will have to work to improve your relationships, do well in your job, perform great in your college etc. Is it important? It is critical. If you want to ask me how critical it is, let me tell you something- when I took CAT in 2004, I had quit my engineering career to prepare for CAT. I knew in my heart that if I didn’t clear CAT I would have a hard time going back to my profession. The pressure killed me in the exam, despite the fact that my All-India ranks in the mocks used to be in top 20. I couldn’t perform in the paper. Come 2005, I started loving my teaching profession, so much so that I wanted to come back to it even if I cracked CAT. I used to teach for 14 hours a day at time (morning 7:00 to evening 9:00) and still come back home and study because I used to feel happy and refreshed. If you can wake up happy every morning to go to your office and come back happy, it means you love your job. Anyways, coming back to my story, where in 2004 I used to score 16- 18 marks in quant and DI, I started scoring 27- 30. A phenomenal jump, not because I was preparing hard- in fact I was studying one-tenth of what I was studying in 2004- but because I felt great about myself. In 2005, when I sat for the CAT paper, the hours passed away smoothly. No screw-up. Same happened in 2008. So if you ask me if you should quit your job or not pay attention to your college exams to prepare for CAT? My answer is that you have killed your chances before you have even started- Total Gadha.

 

This article comes from an old-timer, Software Engineer- commonly called as SE in our forums. When SE came on TG, Dagny and I thought he was a brat but since brats are rare among sweet gadhas of donkeyland we kind of liked it. Well, we still think he is a brat but he is a lovable brat. And an intelligent one too, which you will find out from this article. Young and fresh TGites, please pay attention to this article. You cannot get better fundas than this. And don’t forget to thank SE for his effort!

cat 2009 factorial

cat 2010 factorial

factorial 3

factorial 4

factorial 5

factorial 6

factorial 7

factorial 8

factorial 9

factorial 10 cat 2010

factorial 11 cat 2009

 

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Re: Factorials
by amit jain - Wednesday, 17 June 2009, 05:12 PM
 

I have not read the full article but it looks superb.

Good job !

Re: Factorials
by pushpinder rana - Wednesday, 17 June 2009, 05:59 PM
 

Its T20 version of Number System. Good one.....smile

pushp....

Re: Factorials
by Ayan Ghosh - Wednesday, 17 June 2009, 07:58 PM
 
Thanks yaar...awesome article ....
Almost all the concepts about Facorials reqd for CAT are covered ....
A suggestion....
You could have touched upon Wilson's Theorem....anywayz thats there already in TG....
A ready reckoner for Factorials...
Tnx once again...Ayan
Re: Factorials
by Anandraj T - Wednesday, 17 June 2009, 07:58 PM
  wow.. awesome article.. almost all the shortcuts on factorials are elaborately explained.. Thanks SE.. had a chance to look at it at the time of my Number system preparation.. smile
Re: Factorials
by Harmanpreet Singh - Wednesday, 17 June 2009, 08:26 PM
 

Did I mentioned How much I am impressed with the article? I am software engineer too and to be frank this article had given me impression to share my collection too...So wait there will be another SE post too...pretty soon

Thought to share this share with you...

Mujhe kal par nahin bharosa bankar umeed jo aaye,

jaroorat hai mujhe ek khubsurat aaj ki....

So true words TG, for better future you need to be happy with what you have in present...

Thanks and stay focussed...Cheers!!

 

 

 

Re: Factorials
by Jayesh Parmar - Wednesday, 17 June 2009, 10:20 PM
  I just love the article... I too believe the psychology condition play a big role in your score.

Keep posting such wonderful article.
Re: Factorials
by SHAUNAK A - Wednesday, 17 June 2009, 11:36 PM
  TG Sir

Are these snippets from TG Number System book??

Please do reply ASAP.


@SE Sir
AWESOME Article

Regards:
Shaunak
Re: Factorials
by Total Gadha - Thursday, 18 June 2009, 02:04 AM
  Hi Shaunak,

Yes, nearly 80% of the problems are in TG number system book. Rest of them have been contributed by SE. If you combine the problems in TG's number system book with lessons on CAT CBT Club for your number system preparation, you have an explosive combination. I am doing some selling here, but speaking the complete truth. smile

Total Gadha
Re: Factorials
by SHAUNAK A - Thursday, 18 June 2009, 02:17 AM
  Thanx TG Sir, can i get any printed version of the book??

Regards:
Shaunak
Re: Factorials
by Total Gadha - Thursday, 18 June 2009, 02:20 AM
  Hi Shaunak,

Not yet. You can only get the pdf copy for now. It's too heady for Dagny or me to run around finding a publisher.

Total Gadha
Re: Factorials
by amit kumar - Thursday, 18 June 2009, 02:30 AM
  The article starts with nicely written words by TG

and SE has done a gr8 Job.

Thanks TG and SE,awesome article..

Regards
Amit
Re: Factorials
by vivek ghiria - Thursday, 18 June 2009, 07:19 AM
  Hello Buddy

A nice article and compiled very effectively. Till now I read half of the article. I have one doubts in that, please clarify:

Prob : The quotient obtained by dividing n by 11 and the number 11 are co prime to each other. Find the number of possible value of n if n is less than or equal to 10000.

-> It means that the highest power of 11 contained in the number n is exactly equal to two!!!!!!!!! I am not able to understand this, can you please elucidate.


Also this problem :
Find the number of possible value of n=125*m if n is less than or equal to 1000 and m is not divisible by 5.
When I saw this problem I calculated
m = 1,2,3,4,6,7,8 :D :D total possible number of n = 7
Re: Factorials
by nutty nuts - Thursday, 18 June 2009, 10:39 AM
  Superb ! Great compilation SE !!
I would also like to mention that you have chosen the font style and size very well.
Thanks a lot !
Re: Factorials
by Software Engineer - Thursday, 18 June 2009, 11:01 AM
 
Vivek,

Apologize. sad

Re: Factorials
by shivam mehra - Thursday, 18 June 2009, 11:37 AM
  hey SE!
u hav shown that all the SE's are fantabulous
this article has really improved me a lot
thanks man

~another SE


Re: Factorials
by Software Engineer - Thursday, 18 June 2009, 11:54 AM
  Vivek, TG has corrected the typo . He replaced 11 by 121. smile
Re: Factorials
by CATching CAT - Thursday, 18 June 2009, 03:47 PM
 

only one word

Brilliant !

Re: Factorials
by sumit jamwal - Thursday, 18 June 2009, 05:47 PM
 

smile

hi SE,

simply cool article

SE

sumit

 

Re: Factorials
by vikram singh - Thursday, 18 June 2009, 06:43 PM
  Great Article as always..... could have been posted in 2 parts.... but SE preferred to ctrl+C & ctrl+V in one go..... thanks for the fundas.. smile
Re: Factorials
by sumit jamwal - Friday, 19 June 2009, 09:04 AM
 

cool

hi SE,

as the article is too long if you have a

copy of this article in word or text format then

do tell .

Regards,

Sumit jamwal

Re: Factorials
by Software Engineer - Friday, 19 June 2009, 11:04 AM
  Sumit,

Print out the images. As the size of each image, except last one, is exactly equal to that of the standard A4 size paper; there won't be any trouble. smile

- SE
Re: Factorials
by userdce . - Friday, 19 June 2009, 05:34 PM
    kudos  to TOON DESIGNER
Re: Factorials
by Total Gadha - Friday, 19 June 2009, 08:06 PM
  *bows* thanks user dce smile
Re: Factorials
by sumit jamwal - Friday, 19 June 2009, 10:56 PM
  hi SE,

ya the same i'v been doing till now ..had faced few problems ..sometimes
when article is large you miss the sequence of images ...sad
that's y i was asking about the word format..otherwise the thing would be 
cumbersome..

Regards,
Sumit
Re: Factorials
by Software Engineer - Saturday, 20 June 2009, 01:26 AM
  Sumit,

TG has named the images like this:- SE1.png, SE2.png, SE3.png, ... SE11.png
The exact sequence of the images can be determined through these numbers.

- SE
Re: Factorials
by userdce . - Saturday, 20 June 2009, 02:00 AM
  I was so much carried away by the toon that I forgot to say:



kudos to Software Engineer 
Re: Factorials
by vivek ghiria - Saturday, 20 June 2009, 07:40 AM
  Hi SE

Thanks again for this article. I have gulp it down. However While revising the chapter I noticed type mistake here -

30! when expressed in base 12 ends with K zeros. Find K.

In the solution, The highest power of 3 is greater than that of 2^2, so the highest power of 12 should be that of 2^2. Hence K =13
Re: Factorials
by shivam mehra - Saturday, 20 June 2009, 11:43 AM
 
last 2 digits
concept is not clear
please help 
Re: Factorials
by userdce . - Saturday, 20 June 2009, 05:38 PM
  please TG get any publisher

its too easy in Delhi to find a publisher and also it is better for students to read from book than pdf file
Re: Factorials
by iim freak - Saturday, 20 June 2009, 11:54 PM
 
hi TG sir and SE ..

thanks a ton for posting this article.. its simply awesome specially the successive division to find out the highest power of a number in another num ..its a real time save ..

@SE...i have a small doubt in the below prob

pls guide me how 2 decide that the max power of 11 in the num will be exactly 2 ?




Best Regards
Randeep






Re: Factorials
by Total Gadha - Sunday, 21 June 2009, 12:45 AM
  Hi user dce,

My only gnawing worry this year is to write more and more lessons for my CAT CBT Club. I am okay if I don't sell a single book. But I am not okay if I put the career of my students at stake. Publishers take time. And time is something which is at a premium for me. sad

Total Gadha
Re: Factorials
by Veena Binya - Sunday, 21 June 2009, 12:03 PM
  Hi SE

A lot of effort must have gone into compiling this!

I am wondering if there is a typo in the last question...

This is what that reads in the question..

25! + 26! + 27! + 28! + 30!

When we take out a 25! in common,

shouldn't it be

25! (1 + 26 + 26*27 + 26*27*28 + 26*27*28*29*30)

instead of the

25! (1 + 26 + 26*27 + 26*27*28 + 26*27*28*30)

which appears in your article?

Luckily.. this does not change the answer!


Kindly let me know in case I am wrong.
Re: Factorials
by shikha singh - Sunday, 21 June 2009, 02:13 PM
 

hi SE sir....

thx 4 ur great contribution.....

can u provide sum notes on number system.....smile

Re: Factorials
by Tarun agarwal - Sunday, 21 June 2009, 04:40 PM
 

Hi TG and SE

great is the only word for this article.

one doubt in finding no. of divisors of  any factorial number.What if the no. is too big.for e.g.-no. of divisors of 543!.what approach should we take,since time is a big factor.

please clear the doubt sir.

rgards tarun. 

Re: Factorials
by vivek ghiria - Sunday, 21 June 2009, 05:48 PM
  Hi Tarun

If the finding the highest power of all the prime number less than 543 is the only way to calculate the number of divisor, I am sure no exam will have that big number.
Re: Factorials
by Total Gadha - Sunday, 21 June 2009, 07:10 PM
  Hi Tarun,

I don't even know all the prime numbers in 543!

Total Gadha
Re: Factorials
by Software Engineer - Monday, 22 June 2009, 11:29 AM
 
Vivek, I think TG corrected the typo. Thanks.

Shiavm, Please read the solution of 'Find the last two non zero digits of 11!' and let me know which step is unclear.

RandeepSingh, Let n = Q * 121. Here, Q and 11 are co-prime to each other i.e. Q is not divisible by 11 i.e. Q does not contain 11 as factor in its prime factorization  i.e. the highest power of 11 in the number Q is zero. Therefore, the highest power of 11 in the number n is 2 + 0 = 2. I hope it's clear now.

Veena, Yes you are right. I missed the number 29. sad

Tarun,  Find the total number of divisors of 543!   This problem will consume a lot of time; if you skip it you can solve the whole Quant section within the same time. smile

Shaunak, Shikha  Just say 'SE' not 'SE Sir'.

- SE
Re: Factorials
by debarun chatterjee - Monday, 22 June 2009, 11:19 PM
  that was really awesome
there should be such shortcut sessions on each topic
Re: Factorials
by harish kothari - Tuesday, 23 June 2009, 01:01 AM
  i am always searching for any new posts on this site ...and whenever i find one ..i am more impressed and happy to be a part of TG family......
Re: Factorials
by iim freak - Wednesday, 24 June 2009, 10:10 AM
 

hi SE .. thanks a ton dude for the explaination ..

 

Best Regards

Randeep

Re: Factorials
by Dagny Taggart - Thursday, 25 June 2009, 11:13 AM
  smile
Re: Factorials
by Software Engineer - Thursday, 25 June 2009, 11:49 AM
  smile!    A Smiley Factorial!
Re: Factorials
by Amar Kr Dubedy - Thursday, 25 June 2009, 01:34 PM
  I think there is an easier method to do this stuff.
Lets say we need to find rightmost digit of 23!

Now, rightmost nonzero digit of 10! is 8.

The rightmost nonzero digit of
11 x 12 x 13 x ..... x 20 is 8
and so on.

To find rightmost nonzero digit of 23!,

we just multiply 8^2 x ( 2 x 3 ) which is 4.


For larger numbers, such as 1990!
we can safely say, it should be the unit digit of 8^1990. Since we know the cyclicity of 8 is 4, we can find the unit digit as 4.
Re: Factorials
by Total Gadha - Thursday, 25 June 2009, 01:49 PM
  Hi Amar,

The rightmost nonzero digit of 21 × 22 × 23 × .... × 29 × 30 is not 8. Neither is that of 31 × 32 × 33 × .... × 39 × 40

Total Gadha
Re: Factorials
by vishal shah - Thursday, 25 June 2009, 03:38 PM
  A superb article...Thanks for this SE...smile
Re: Factorials
by nidhin john - Thursday, 25 June 2009, 08:35 PM
  hi SE,
        The article is superb as it gives a lot of inputs and fundas.Considering Numbers as one of the most difficult  part  in CAT quant section,these will be surely useful.
Here I have some more interesting number questions dealing with factors
===================================================================

Let a number be denoted as N,N=a*b. where a ,b are prime numbers

 Then number of factors N has =(m+1)(n+1).

Sum of factors of N, S={a^(m+1)-1/a-1}* ={b^(n+1)-1/b-1}

==========================================================

Some previous CAT questions just from the above number property......


*Find the number of factors 24 has.

 

Explanation:-

24 can be written as 2^3*3.

We know that if N=a^m*b^n. Then the number of factors of N is (n+1)*(m+1)

Here we have, number of factors= (3+1) (1+1) = 8.

 

*Find the least number with no. of factors as 24.

 

Explanation:-

A number can have 24 factors,

 If  its in the form, N=a^23,

                                 or  a*b^11

                                  or a^2*b^7 

                                  or  a^3*b^5

                                   or a*b*c^5

                                   or a*b*c*d^2 .

 

Here if we want to find the smallest number we need to fit in the smallest prime numbers as far as possible.

So better if you put the last choice,  a*b*c*d^2, also put d=2,c=3,b=5,a=7

    So the number becomes 2^2*3*5*7 ie

                                            4*3*5*7= 420.

So this is the smallest number with 24 factors.

If we put any other forms of the number, the resulting number will be higher than 420.

 

*A number has unequal prime factors and the total number of factors is 4.

  The sum of the factors without 1 and the number itself is 30. Find the smallest number with this characteristics.

 

Explanation:-

 The number with 4 factors can be with the form a^3 or a*b.

  Since its given that number has unequal prime factors, it can take the form of  a*b.

  Then the factors are 1, a, b, ab where ab is the number itself.

The condition given is a+b=30, where a and b are prime numbers.

 a,b can be 7 & 23

                   11 & 19

                    13  & 17.

Then the smallest number is 7*23 = 161.


regards,


Nidhin

Re: Factorials
by B a D r I Life Is sO bEauTifUL - Friday, 26 June 2009, 01:43 PM
  What you said is 200% correct.. I faced this situation last year...
Re: Factorials
by srinivasan ravi - Saturday, 27 June 2009, 12:27 PM
  hi all,
pls help me..thanks..
1. What is the number of ending zeros in 1^1 * 2^2 * 3^3........ 99^99*
100^100.
a. 1050
b. 1300
c. 1250
d. None of these.
Re: Factorials
by sumit jamwal - Tuesday, 30 June 2009, 05:22 PM
  1100..option d smile
Re: Factorials
by Software Engineer - Tuesday, 30 June 2009, 05:52 PM
  Srinivasan, Sumit

Let's calculate the highest power of 5 in the given expression by writing only multiples of 5
5^5 * 10^10 * 15^15 * 20^20 * 25^25 * ... * 100^100

Each of the term in the expression contains at least one power of 5.
5^5 * (5*2)^10 * (5*3)^15 * (5*4)^20 * (5*5)^25 * ... * (5*10)^50 * ... * (5*20)^100

The powers of such terms are 5, 10, 15, 20, ... 100. The sum of the powers = (10) (5+100) = 1050

After removing exactly one power of 5 from each term of the above expression, there exists some terms which still contain some power of 5. Initially the terms were 25, 50, 75 and 100. After removing 5^1 from each of these terms
5^25 * 10^50 * 15^75 * 20^100
The highest power of 5 in this expression = 25 + 50 + 75 + 100 = 250

Hence, the total number of ending zeroes = 1050 + 250 = 1300.

- SE
Re: Factorials
by avinanda dutta - Wednesday, 1 July 2009, 01:07 AM
  Ans : b. 1300
for calculating no of zeros, u need to calculate the no of 5's....coz zeros come as the product of 5's n 2's and thr r more 2's than 5's.
So, it is :
(5^5 * 5^10 * 5^15 ....5^100)*(5^25 * 5^50...5^100)
the last bracket is for the multiples of 25's
it gives 5^(1050 + 250) = 5^1300
So, the no of zeros = 1300
Re: Factorials
by Ankita Chowdhury - Tuesday, 7 July 2009, 09:00 PM
 

Hi TG/SE,

Please explain me the diff between the questions:

1.How many numbers les than or equal to 1000 are divisible by 125?
2.Find the number of possible values of n = 125*m if n is less than or equal to 1000 and m is not divisible by 5.

TO me, the 2nd question is also about finding possible values of n =< 1000 such that it is divisible by 125.

Waiting for ur reply.

Ankita. 

Re: Factorials
by Software Engineer - Wednesday, 8 July 2009, 01:41 AM
  Ankita,

Q1 means that find all the numbers which are less than or equal to 1000 and the highest power of 5 in each of these numbers is greater than or equal to 3.

Q2 means that find all the numbers which are less than or equal to 1000 and the highest power of 5 in each of these numbers is exactly equal to 3.

Solution1 includes 625 but Solution2 does not as the highest power of 5 in 625 is 4.

- SE
Re: Factorials
by Ankita Chowdhury - Wednesday, 8 July 2009, 08:13 AM
  Thnx a ton SE for a very clear explanation! smile
Re: Factorials
by sumit jamwal - Thursday, 9 July 2009, 05:07 PM
 

hi SE,

yup!! you are right ..i did not see 100^100  sad

so 1100 + 200 (for 100^100) ==1300  smile

 

Regards,

Sumit Jamwal

Re: Factorials
by AsHwIn Drmz - Friday, 10 July 2009, 12:47 PM
 

Mind blowin .....article.I Always...keep looking for new articles in TG,and when i find anythin new i get excited bcos i m always sure that the article will be ausome without any doubt.

Thanx a ton .... smile

Re: Factorials
by Puneet Aggarwal - Saturday, 11 July 2009, 07:33 PM
  This is the best article I have read so far......cleared lot of things....thanks a lot
Re: Factorials
by AsHwIn Drmz - Monday, 13 July 2009, 11:17 AM
 

hi Tg,

This is regarding the PDf Number system and geometry PDF books.Are the PDFs in printable format or will they be protected?

Re: Factorials
by Total Gadha - Monday, 13 July 2009, 01:41 PM
  Hi Ashwin,

They will be printable.

Total Gadha
Re: Factorials
by Gowtham Muthukkumaran Thirunavukkarasu - Wednesday, 15 July 2009, 12:21 AM
  I would like to add to the second funda
1*2*3*4 will end in 24
6*7*8*9 will end in 24
1471*1472*1473*1474 will end in 24
.....
...... and so on, 24*24 will end in 576


like in case of 4&6 where
4*4 ends in 6
4*6 ends in 4
6*6 ends in 6

24*24 ends in 76
24*76 ends in 24
76*76 ends in 76
Re: Factorials
by Gowtham Muthukkumaran Thirunavukkarasu - Wednesday, 15 July 2009, 01:17 AM
  Juz awesome! I completely respect and appreciate your effort. Worth ur and my time. Thank you once again and keep up the good work. Thank u TG team. Is there a way to print these? If so please make it known.
Re: Factorials
by krishna reddy - Wednesday, 15 July 2009, 12:14 PM
  thamk you yar
Re: Factorials
by sankalp bansal - Friday, 17 July 2009, 12:44 AM
  Hi SE



I think this question doesn't say whether there would be remainders or not after division by 121.....therefore for every number we have to take 120 more numbers considering remainders with each and every one...isn't it??
Re: Factorials
by Software Engineer - Friday, 17 July 2009, 09:12 AM
  Sankalp,

n = 121 * Q.
As n is always divisible by 121, the remainder when any possible value of n divided by 121 is always 0.

- SE
Queries regarding CAT CBT Club
by Umang Mathur - Friday, 17 July 2009, 11:43 AM
  Dear TG,
             This is an awesome article. 
I am agog to see such concepts, beautifully explained. I want to be a member of the CAT CBT Club and want you to throw some light on some of my queries. cool

a. Will I get the access to the various TG e-books, like e-book on Number System, Geometry etc?

b. TG, apparently, seems to be a place which is more for Maths. I was unable to find some good article on English & DI. Is it so, that CBT will give an insight to these?

Regards evil
Umang


Re: Factorials
by sankalp bansal - Friday, 17 July 2009, 02:56 PM
  Hi SE
excuse my prying...bt its just tht i cudn't see whr in the question its given...that n is divisible by 121....still point understood...thanx a lot...
Regards
Sankalp
Re: Factorials
by Abhiseka Sahu - Saturday, 18 July 2009, 04:49 PM
  Thanks SE,
For this wonderful article.
Re: Factorials
by gaurav kaushal - Sunday, 19 July 2009, 10:22 PM
  hi sir ,

find the number of divisors in 15!..
i dint get the last line of this question ...
the number of divisors of 15 ! = 12  *  7  *  4  *  3  *  2  *  2 = 4032

cud u plz help me out ?
thanks !!!
Re: Factorials
by Sanjay J - Monday, 20 July 2009, 04:34 PM
 

Hi ,

 

I have a doubt in qustion - "How many nos less than or equal to 9000 are a multiple of 49"

What i understand is we have to count no. of seven squares.

9000 -- no of sevens 1285 + 183 + 26 + 3 = 1497

Therefore no. of 7 squares woud be 1497/2 = 743

So ans should be 743.

If not please let me know where am i wrong

 

Thanks!

 

Re: Factorials
by Software Engineer - Tuesday, 21 July 2009, 10:29 AM
  Gaurav,

4! = 2^3 * 3^1
Total divisors of 4! = (3+1) * (1+1) = 8.

Please refer, Divisor Lesson

- SE
Re: Factorials
by Software Engineer - Tuesday, 21 July 2009, 10:40 AM
  Sanjay,

The highest power of 7 in 9000 = 1285 + 183 + ...

There are 1285 numbers less than or equal to 9000 which are divisible by 7.
There are 183 numbers less than or equal to 9000 which are divisible by 49.
and so on.

Please read "The highest power of 2 in 20! is 33. What does it mean?"

"9000 -- no of sevens 1285 + 183 + 26 + 3 = 1497
Therefore no. of 7 squares woud be 1497/2 = 743"
-- What you are counting is the highest power of 49 in 9000!.

- SE
Re: Factorials
by Varun Khanna - Monday, 3 August 2009, 04:49 PM
 

CAT doesnt require this level of number system fundas....

SE,you should have quoted only relevant concepts.

And I totally disagree with your article that qutting job and preparing for CAT is a mistake.Completely depends on an indiviudual what lifestyle does he carry after qutting the job.

Like maths,psychology can never be generalised my friend.Many aspirants loose the fight and give up the CAT preparation under enormous job pressure .Many of those guys are deserving but the work envoirnment around them is not favourable enough to keep up their enthu for the preparation.The problems in the work envoirnment could be because of irrational reasons which the aspirant may not be able to identify.

In short,to crack CAT,whether you quit or continue working,one should carry his lifestlye such that he remains positive and can mentally convince himself that he is a winner and could achieve a good percentile.....

 

Re: Factorials
by krunal dhamesha - Friday, 7 August 2009, 11:17 AM
  the explanation of d sum 25!+26!............+30! seems faulty...............u miss one summation of 29! wen u r summin up last tw0 digits instead u took 30...........................nd for simple answer one can go for nly last digits...........nd once found the last digit of 25!........which is 4 jus multiply it wid 6 so it vl gv u last nonzero digit of 26!....nd so on..........nd by doin dat...............u can sum up all the last digits nd see wether it is generatin a new zero or nt..............it wont give a zero......................bt the quality of ur article is awesome........................pls keep posting this kinds of article...........
Re: Factorials
by Total Gadha - Friday, 7 August 2009, 07:46 PM
  Hi Krunal,

The question does not contain 29! It has been done on purpose.

Total Gadha
Re: Factorials
by krunal dhamesha - Saturday, 8 August 2009, 12:24 AM
  ohhhh.........yeah...........dis is d kind of mistakes i alwaz do..............nt readin d que. properly..............thanx for drawin my attention....................
Re: Factorials
by Nitin Negi - Tuesday, 11 August 2009, 01:56 AM
  Hi SE,

This is really a awesome article, not only did this cleared the fundamentals, but made them so strong that I found a new techique of solving some of the problems offered by you.

Que     : Find the rightmost non zero digit of 34!
My Way: 34! has 7 powers of 5 and 32 powers of 2. if we remove 7 power of 5 and 2 ( which contributes to Zero) and write down 34! as product of all Prime Factors

34! = (2 pow 25 )* (3 pow 15 )*( 7 pow 4) * (11 pow 3) *.............* (31 pow 1)

Note : 1) Power of all the prime factors found using same division method
2) Power of 2 is 25, because 7 are used by 5.

now using cyclicity property unit digit of all the terms can be found and multiplying all the unit digits u get your answer....
Any Feedback???

Re: Factorials
by Software Engineer - Tuesday, 11 August 2009, 09:50 AM
  Nitin,

The method described is correct but it consumes a lot of time. It's really difficult to find the rightmost non-zero digit of 121!, by using this method. But if you use the method described in the article, you can calculate the rightmost non-zero digit of 121! within 50 seconds.

If anything is posted on the homepage of www.totalgadha.com, that thing is always awesome. smile

- SE
Re: Factorials
by Nitin Negi - Tuesday, 11 August 2009, 03:28 PM
  Hi SE,

You are totally correct, its problematic for higher no.....but i was happy that i found a now way and hence wanted to tribute it to clarity of the concepts explained by you...

thanks once again

Cheers.
Re: Factorials
by karan krishan - Thursday, 13 August 2009, 06:10 PM
  Great ..
thanx
Re: Factorials
by saitu gupta - Saturday, 15 August 2009, 12:50 PM
 

hi SE

i hve the foln. doubts pls  n e 1, try 2 fnd d soln

 

rem when 55555555................93 times / by 98

12341234.....89 times / by 19

find last 3 digits of (12363) ^ 1998

Re: Factorials
by Siva Thanu - Saturday, 22 August 2009, 12:32 AM
  Hi TG and SE,
I am a neophyte to TG..Amazing articles..Expecting more from you..smile Thanks a lot for ur Articles...
Re: Factorials
by Abhishek Panda - Saturday, 29 August 2009, 09:28 PM
  This has been one helluva article....Thanks for this awesome contribution....Seriously the best in class!!!
Re: Factorials
by Umashankar k - Sunday, 30 August 2009, 02:02 PM
  Find the number of possible value of n=125*m if n is less than or equal to 1000 and m is not divisible by 5?

for this question how about m=0,-1,-2,-3....
Re: Factorials
by vikas sharma - Thursday, 3 September 2009, 03:43 PM
 

Hi Tg sir

pls tell me sol of following ques.

How many positive integers less than 100 can be written as the sum of 9 consecutive positive integers‌

Re: Factorials
by vishal shah - Friday, 4 September 2009, 07:21 PM
 

Hi Vikas,

I have taken a very manual approach to tis problem. But still I took less than a minute to solve it.

Do the addition of first 9 natural numbersi.e. n(n+1)/2...here n=9...so sum is 45.

Next take 2-10...exclude 1 from the above sum & add 10..so sum is 54...

next 3-11..exclude 1 nd 2 from 45 nd add 10 nd 11...so sum is 42+10+11=63...

we get a pattern of numbers with a common difference of 9...so less than 100 starting from 45 we have 7 such numbers...45,54,63,72,81,90,99...

Let me know if you have any doubts...

Re: Factorials
by Mohit Bhambri - Saturday, 5 September 2009, 07:12 AM
 
Engineer ki lingo mein bole to "Fod daala" software engineer..:D..mast samajh aaya factorials..maza aa gaya smile
Re: Factorials
by Kunal Parekh - Sunday, 6 September 2009, 10:47 PM
 

Dear SE,

Thanx a lot for your useful article..........

Kunal

Re: Factorials
by Netra Mehta - Monday, 7 September 2009, 02:37 AM
  Plz solve dis problem....

Find the no. of consecutive zeroes at the end of S where S= 5^1 + 10^2 + 15^3 +...100^20

A. 245
B. 145
C.160
D.147
E.210
Re: Factorials
by Janmejaya Panda - Tuesday, 8 September 2009, 09:30 AM
  Wonderful Article indeed. Thanks software engineer.

Re: Factorials
by vikas sharma - Tuesday, 8 September 2009, 05:08 PM
 

Thanks alot vishal

 

Re: Factorials
by tulasi ram - Wednesday, 9 September 2009, 10:03 PM
  fundoo concepts SE,
Reallly superb., evil
Re: Factorials
by cute devil - Sunday, 13 September 2009, 11:14 PM
  Great sharing SE....u deserve a high respect in this community of sharing knowledge...!!
thankx a zillion !!!!
br,
CD..
Re: Factorials
by Nishant A - Friday, 18 September 2009, 06:31 PM
  kudos to u SE and TG....
every concept is covered here ...
warm regards
Nishant
Re: Factorials
by Vipul Gupta - Tuesday, 22 September 2009, 07:45 PM
  can smeone tell me the last two digits of 36!, kindly give the solution also.
Re: Factorials
by ROHIT K - Tuesday, 22 September 2009, 09:26 PM
  Hi vipul,

last two digits of 36!

by observation u can find that 36! has more than 2 zeroes @ the end.

Checking highest power of 5 in 36!

36/5 = 7
7/5 = 1

Total number of zeroes @ end = 7+1 =8

Hence, the last two digits are 00

Rohit

P.S. Is this the question or the last two non-zero digit the question?..plz confirm..thankx

Re: Factorials
by Vipul Gupta - Thursday, 24 September 2009, 05:59 PM
  can smeone tell me the last two digits of 36! excluding the zeros, kindly give the solution also.
Re: Factorials
by abhinav prakash - Friday, 25 September 2009, 01:18 PM
 

i guess the answer is 52:

after removing all the 5s the last two digit comes outta be 12.

x12 = 2^8 X xab.

last digit using fundas in the article comes out to be 2 and not 7.

Now, x12 = x56 X xa2

So, 6a + 1 = 1

Only valid for a = 5.

Thus its 52

Re: Factorials
by Milind Sharma - Tuesday, 6 October 2009, 10:58 AM
  Hi TG, refering to your introductory paragraph, well even I am a Software Engineer smile and struggling to crack CAT for since 2008. You said one has to have a healthy psychology and should love his job...well what if i don't love my job at the first place?? How can i still mantain this healthy psychology? should i quit my job eventually?
Re: Factorials
by mayuri joshi - Thursday, 29 October 2009, 09:00 AM
  hi se, tg
in the same sum that u are discussing even if the question doesnt contain 29! when u split up the last term into 30*29*28*27*26 -- u will still require the 29 term since it is 30!


Re: Factorials
by s ram - Thursday, 29 October 2009, 10:19 PM
  Dhamaal of an article

Thanks SE fr sharing the wealth that you have with all of us.

smile smile
Re: Factorials
by Kishan Kumar - Wednesday, 18 November 2009, 01:55 AM
 
HI engineer SAHAB

I HAVE ONE QUERY IN YOUR EXAMPLE OF LAST TWO NON-ZERO DIGITS IN 11!

IS THE ANSWER 72 OR 68

IF 68 THEN I AM TOTALLY CONFUSED ABOUT HOW?


PLEASE DO REPLY
Re: Factorials
by amar goswami - Thursday, 19 November 2009, 07:15 PM
  Hi Kishan,

    Let us not disturb BIG GUNS for ur question and allow me to take this opportunity to clarify your doubt here. smile

What we are doing in the first part is that we are dividing the number by 5^n where n = highest power of 5 and calculating the last two digits of the number. Now in the second part we are dividing the number, thus obtained, by 2^n to calculate the last two non zero digits. Because to find last non zero digit we have to divide the whole number by 10^n and here it is being done in parts, first by 5^n and then by 2^n.

Hope it helps. smile
Rgds
Amar
Re: Factorials
by amrit jajodia - Wednesday, 25 November 2009, 07:10 PM
  Hi SE,

In the last question when 30 factorial is considered 29 is not taken into account. that is it should be 26*27*28*29*30.

But the ans wont get effected.

Anyways was an awesome article.

Thanks
Amrit

Re: Factorials
by Aspirend Achieve - Thursday, 22 April 2010, 12:35 PM
 

Hi SE,

Article is extrememly superb....smile

Re: Factorials
by Kushagra Saxena - Wednesday, 5 May 2010, 11:16 PM
  amazing........cleared most doubts...
thanks SE
http://totalgadha.com/pix/s/welldone.gif
Re: Factorials
by kamlesh kholiya - Monday, 10 May 2010, 06:21 PM
  fine article i think everything covered about factorials but is there in any proof i mean mathematical proof or it is a pure observation on smaller number anfd then generalisation
Re: Factorials
by Kartick M - Wednesday, 12 May 2010, 12:47 AM
  the topic seems simple but the kind of funda discussed here, its juss awesome and shows hoe much I under estimated this important topic! GREAT article! Thankx SE!!!!
Re: Factorials
by xlr pmir - Monday, 17 May 2010, 11:22 AM
  thanx ,smileth two articlson rmaindrs and factorials, i hav rad thm all full and vry carfully, gaind a lot rally 
Re: Factorials
by Subhash Medhi - Friday, 21 May 2010, 08:19 PM
  Dear SE sir,
                I have a doubt in your last but one question. I am including the relevant portion of the question below:

x76 = x16 * ab
xx6 = xx6 * b ,b= 6
x76 = x16 * a6
(3 + 6a) + (6) = x7   (6*6 = 36, Carry = 3, 6*a = 6a, Therefore, (3 + 6a). 1*6 = 6                                 Therefore,(6))
6a + 9 = x7
6a = (x-1)8               (Don't 7 - 9 = -2.Do x7 -9 = (x-1)8 )
a = 3 or 8. Therefore, the last two non-zero digits of 24! are 36.

How did you get a = 3 ? Kindly explain.

Regards,
Subhash
Re: Factorials
by Software Engineer - Tuesday, 25 May 2010, 11:41 AM
  Subhash Medhi,

ab= 36 or 86

last two digits must be divisible by 4 (Funda#6)

86 is not divisible by 4

so, ab=36

- SE
Re: Factorials
by Nitin Kumar - Tuesday, 1 June 2010, 11:20 PM
 

Can any one please tell me what is the sum of:

S = 1!+2!+3!+4!+...........+n!

Re: Factorials
by tarkik dave - Sunday, 20 June 2010, 06:25 PM
 

10000! = (100!)K × P, where P and K are integers. What can be the maximum value of K?

102

103

104

105

Re: Factorials
by Abhirup DebRay - Sunday, 20 June 2010, 07:02 PM
  i think its 103
pls confirm
Re: Factorials
by deepika guliani - Saturday, 7 August 2010, 12:01 PM
  GREATTTTTTTTTTT Article.Thanks indeed.
KUDOS!!!! big grin
Re: Factorials
by dharmendra mishra - Friday, 13 August 2010, 11:02 AM
  A very useful article.Thanks for the post.But I have a doubt.U say that (for eg) for calculating multiples of 125 less than or equal to any given no.We have to find power of 5 upto 3 places in the expansion.Will the no. of multiples of 54, 55,56.... not be the multiple of 125.
             Please clarify my doubt as soon as possible.And thanks once again for such a beautiful post.







Re: Factorials
by kaushal roonwal - Saturday, 4 September 2010, 03:25 PM
 

The article is exceptional... It'll help a lot...

Thanks smile

Re: Factorials
by hardik gogri - Sunday, 5 September 2010, 09:39 PM
  Thanks a lot SE
Re: Factorials
by sunny kumar - Sunday, 5 September 2010, 09:48 PM
  can you please share the link of these ebooks...
Re: Factorials
by sharad mishra - Saturday, 18 September 2010, 12:00 AM
  smile
kudos hats off whatever i say is less ,read many articles in gadha land but i must say this one looks like u had given time writing this article
great source for factorial
i just really loved it
again thanks se bhai
by the way what are u doing nowadays
Re: Factorials
by swayam aahuja - Tuesday, 28 September 2010, 12:21 AM
 

Hi SE in your last sum---

how many zeros are present at the end of 25!+26!+27!+28!+30!?

25!+26!+27!+28!+30!=25!(1+26+26*27+26*27*28+26*27*28*29*30) there is a typing mistake there in the explanation.I am really grateful to you for sharing the concept with us....thanks allot.

Re: Factorials
by Subrahmanyam Kancherla - Sunday, 7 November 2010, 10:54 PM
  Sir,
For the question "Find the smallest possible integer n such that n! is multiple of 102009."
8045! will have 2008 trailing zeroes and 8050! will have 2010 zeroes.But 10*8045! should have 2009 trailing zeroes .So answer should be 10*8045! rather than 8050 !(As question asks for smallest possible integer).....Please correct me if I am wrong...
Re: Factorials
by Subrahmanyam Kancherla - Thursday, 11 November 2010, 09:29 PM
  Sorry Sir..above query is illogical....
Re: Factorials
by Saurabh Koshti - Wednesday, 30 March 2011, 01:05 PM
  I m so amazed by this article i can't tell u !!!!!!!!!!  all i ever knew about
factorial  was to keep multiplying till u reach 1 , but its whole new world out here!!

Thank you  "SE Sir".. ***You made World a better place***
Re: Factorials
by prabhat taneja - Wednesday, 17 October 2012, 05:14 PM
  Hi SE,

I would like to thank you and TG team for this effort. Such awe inspiring posts. Thanks a lott..
Sad, I discovered these only a few days back and cannot go through all of them before my CAT. sad

While going through this post, I noticed a discrepancy in calculation last 2 digits of 25!

25! is 15511210043330985984000000

11! is 39916800

Please enlighten me on the same.
Re: Factorials
by GAUTHAM thomas - Thursday, 22 November 2012, 10:08 PM
  software enginner saab .. you are god!!!
i really wish you guys come to chennai and south india too and open TG centers here sad Brilliant article sirr!!!!!
Re: Factorials
by DeeScript .in - Wednesday, 12 June 2013, 07:56 AM
  the higest power of 2 contained in 20! is 33.....how is that possible????

Re: Factorials
by Dipen Jain - Wednesday, 21 May 2014, 08:02 AM
 

hi.. awesome article... got to know many new concepts and many doubts have also been cleared now thanks to ur article...

i have a little doubt if u can kindly address to...
there is one more method to calculate the last non zero digit of a factorial i.e
if we want to calculate right most non zero digit of n!
we use following steps:
write n=5*A+B (A,B are integers)
rightmost non-zero digit of n! given by R(n)=unit digit of (2^A)*R(A!)*R(B!)

now by using this method when i am trying to calculate for 50! i am getting non-zero digit as 2(50=5*10+0, R(50!)=2^10*R(10)*R(0)=4*8*1=32
whereas calculating by ur method i am getting 6 sad
50=10+2
40*8
(8*5+0)(8*1+0)=6*6=6
x6=2^12*r=6*r
r=6
and the value of 50!=30414093201713378043612608166064768844377641568960512000000000000.
so the right most non zero digit has to be 2.
please help me with this..

regards,
an ardent fan of ur work.
Re: Factorials
by dhanu shan - Thursday, 26 June 2014, 01:22 PM
  I couldn't understnd how to find the value of 'a' in last 2 non-zero digits of 25! . Help me wid tt .?
Re: Factorials
by amber gupta - Monday, 13 October 2014, 05:44 PM
  Simply awesome..
hats off to you guys cool cool cool
Re: Factorials
by vaibhav jain - Wednesday, 15 October 2014, 11:09 AM
  Can u please explain the concept used to solve this question.

the quotient obtained when n is divided by 121 and 11 are co-prime numbers?