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by shankar.ashwin » Sat Nov 19, 2011 12:21 am
This will have 45/5 + 9/5 = 10 zeros IMO. After that we would be having a borrow during subtraction and hence after 10 zeros you will have a non-zero term
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by vishal.pathak » Sat Nov 19, 2011 12:34 am
vishal chugh wrote:what is the no. of zeros at the end of the expression
2500!- 45!?
the factors of 10 are 5 and 2. Since every alternate number is even so the number of 2's in any factorial will be less than the number of 5's. This means that the number of 0's in any factorial will be governed by the number of 5's in it.

2500! - 45! will have as many 0's in the end as the number of zeros in the smaller number, barring the case in which the digits immediately preceding those 0's is the same in the 2 numbers. We can reject this case as it is clear that 2500! has a lot many 0's than 45!

No of 0's in 45! = no of 5's in it = 10. IMO 10
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by shankar.ashwin » Sat Nov 19, 2011 12:36 am
vishal.pathak wrote: No of 0's in 45! = no of 5's in it = 10. IMO 10
Yeah, thats what I have done. Number of 0's in a factorial is nothing but number of factors of 5 in it.
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by vishal chugh » Sat Nov 19, 2011 1:42 am
take 45! common, you r left with
45!(10K - 1); mean this no. has only zeros which are present in 45!; further 10K-1 is not a multiple pf 10; mean it has 0 no. of zeros at the end
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by rattan123 » Sun Nov 20, 2011 4:28 am
take 45! common, you are left with
45!(10K - 1); mean this no. has only zeros which are present in 45!; further 10K-1 is not a multiple pf 10; mean it has 0 no. of zeros at the end
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