The Number of Zeros at the End of 1000! (1000 Factorial)
How Many Zeros Are at the End of 1000 Factorial? Find Out Here!
We examine the factors of 1000! to see which can be used to produce multiples of 10.
For 1000!, each multiple of 5 - taken together with some even number - contributes a multiple of 10. Each multiple of 25 - taken together with some multiple of 4 - contributes two multiples of 10. Each multiple of 125 - taken together with some multiple of 8 - contributes three multiples of 10.
We can count each multiple of 5 in 1000! once. Then we can count each multiple of 25 in 1000! once in order to altogether count each multiple of 25 twice. Then we can count each multiple of 125 in 1000! once in order to altogether count each multiple of 125 thrice.
There are 1000/5 = 200 multiples of 5 in 1000!
There are 1000/25 = 40 multiples of 25 in 1000!
There are 1000/125 = 8 multiples of 125 in 1000!
Adding them together, we have 200+40+8 = 248 zeros at the end of 1000!.
Published by G. Stolyarov II
G. Stolyarov II is a science fiction novelist, independent essayist, poet, amateur mathematician, composer, author, and actuary. View profile
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6 Comments
Post a Comment249 is correct, for sure.
Number 652 is 5x5x5x5, this 5 is not included in the sum 248.
Author should add this line:
There are 1000/652 = 1 multiple of 625 in 1000!
249??
mohit is correct
Hello,
I will throw some light on your question, hope it will help.
We are dividing by 5 because by multiplying 5 with 2, it will be 10 that means one zero.
Ex:
5! = 1*2*3*4*5 = 10*12=120 ( one zero)
10! = 1*2*3*4*5*6*7*8*9*10 = 10*10*3*4*6*7*8*9 = 100*72*56*9 ( 2 zeros)
In the same manner when the numbers are being multiplied, 125 will give 2 zero, one have been counted already in case of 5 and second we are counting, dividing the given number by 25.
in the same manner 625 will give 3 zeros, two have been counted already one in case of 5 and the other in case of 25, that's why one zero is being counted for 625.
And we need to take the integer part only(ex: if output is 2.34 then the correct ans is 2)
so for 1000!, number of zeros are:
1000/5 + 1000/25 + 1000/125 + 1000/625 =
200 + 40 + 8 + 1 ( ~1.6) = 249
why use the 5 for denominator?? Thanks MAry
Note: THE ABOVE ANSWER IS INCOMPLETE.
We also need to consider multiples of 625 in 1000! -- of which there is one. 625 = 5^4. Together with a multiple of 2^4 = 16, a multiple of 625 will result in 4 multiples of 10.
So to complete the answer, we need to add another zero to the 248 previously obtained -- in order to get a total of 249 ZEROS.
My apologies for the initially incomplete answer.