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number properties

by vscid » Mon Apr 12, 2010 8:17 pm
If 60! is written out as an integer, with how many consecutive 0's will that integer end?

1]6
2]12
3]14
4]42
5]56

What is a quick way to solve this?
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by liferocks » Mon Apr 12, 2010 8:41 pm
number of 5s in 60! will be the number of 0s as 5x2 give 10 and 2 will outnumber 5 as it is present in every alternate number
number of 5s in 60! is
60/5+60/25=12+2=14
hence ans is [spoiler]option 3-->14[/spoiler]
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by lkm » Mon Apr 12, 2010 8:48 pm
vscid wrote:If 60! is written out as an integer, with how many consecutive 0's will that integer end?

1]6
2]12
3]14
4]42
5]56

What is a quick way to solve this?
The quickest way to solve this is to identify the number of 5's in the 60!

So, five will appear in the following numbers:-

5
10
15
20
25 (5x5) --> There will 2 fives in 25
30
35
40
45
50 (5x5x2) ---> Also, there will be 2 fives in 50
55
60

So, in total there are 14 fives. Hence, there will be 14 zeros at the end if we expand 60!.

Ans: 14
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by koolk86 » Tue Apr 13, 2010 11:46 pm
hey guys i have found some easier method to solve this:

To find the number of zeros in n! one can use this formula

n/5 + n/5^2 + ...............+ n/5^k

till 5^k < n

i this question to find no of zeros in 60! we can use

60/5 + 60/25 = 12 + 2 = 14

( we take it till 25= 5^2 since 5^3 will be > than 60 )

Sorry just saw the post above the method has already been explained


Hope this helps ...
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