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?
number properties
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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]
number of 5s in 60! is
60/5+60/25=12+2=14
hence ans is [spoiler]option 3-->14[/spoiler]
The quickest way to solve this is to identify the number of 5's in the 60!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?
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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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 ...
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 ...