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Shivam Bhagat Grade: 9
        

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

6 years ago

Answers : (2)

Surya Anuraag Duvvuri
39 Points
										

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.

6 years ago
SAGAR SINGH - IIT DELHI
879 Points
										

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.


 





































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6 years ago
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