Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.

Use Coupon: CART20 and get 20% off on all online Study Material

Total Price: Rs.

There are no items in this cart.
Continue Shopping

(1*2*3_________*200) is a number. how many Zeros will be their in the end.???? pls explain

(1*2*3_________*200) is a number. how many Zeros will be their in the end.???? pls explain


2 Answers

Surya Anuraag Duvvuri
39 Points
10 years ago

Dear Shivam Bhagat,

         Your Question is (1*2*3*............................*200) ends with ........................ zeros.

Ans: It can be solved using Number Theory ,

     So,you like to find no. of zeros at the end of   (1*2*3*............................*200) i.e., 200!

                                                                  = [200/5]+[200/25]+[200/125] =40+8+1 =49

 Therefore at end of 200! there are 49 zeros.

Example. Let us find the no. of zeros  at the  end of X!

   =[X/5]+[X/25]+[X/125]+.......................                                              ( Denominators are the increasing powers of 5 )

                                                                                                                                                     Where  [.] denotes the Integral value.

 All The Best.

Surya Anuraag, Your Freind.

879 Points
10 years ago

Dear student,

Note that 200! = 200 x 199 x 198 x … x 3 x 2 x 1, by definition.

Finding how many zeros 200! ends with is the same as finding how many consecutive times 200! can be divided by ten without leaving a remainder.

To this end, let us try to compute how many 5′s are in the prime factorization of 200!.

Each of 5, 10, 15, …, 200 has a factor of 5 in it, so that gives 40 fives in the prime factorization of 200!.

In addition, each of 25, 50, 75, …, 200 has a second factor of 5 that we haven’t counted yet, which gives 8 more fives in the prime factorization of 200!.

Finally, 125 has a third factor of 5, which gives one additional factor of 5 in the prime factorization of 200!.

So 200! = 5^49 * (some number not divisible by 5).

It should be evident that 200! has many, many more copies of 2 in its prime factorization than it has copies of 5. (I won’t compute exactly how many copies of 2 it has, because it’s not necessary.)

So we can write 200! = 2^49 * 5^49 * (some number not divisible by 5).

Thus, we can divide 200! by 10 only forty-nine times; after that we get a number that isn’t divisible by 5, hence not by 10 either.

So 200! ends in forty-nine zeros.


Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.

All the best.

Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.

Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar  respectively : Click here to download the toolbar..


Askiitians Expert

Sagar Singh

B.Tech, IIT Delhi

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy


Get your questions answered by the expert for free