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(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

Grade:9

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.

SAGAR SINGH - IIT DELHI
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.

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Sagar Singh

B.Tech, IIT Delhi

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