 ×     #### Thank you for registering.

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

Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

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

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
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

```
9 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.
```
9 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.

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.

Sagar Singh
B.Tech, IIT Delhi
sagarsingh24.iitd@gmail.com

```
9 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Magical Mathematics[Interesting Approach]

View all Questions »  ### Course Features

• 731 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution  ### Course Features

• 731 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions