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bmwhype2
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what is the largest integer k such that 15! is divisible by 20 Dec 2007, 13:18
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what is the largest integer k such that 15! is divisible by 10^k?


what is the largest integer k such that 20! is divisible by 10^k?



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what is the largest integer k such that 15! is divisible by 20 Dec 2007, 14:06
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what is the largest integer k such that 15! is divisible by 10^k?


what is the largest integer k such that 20! is divisible by 10^k?
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These problems are typically worded as "how many zeros at the end of x! ?

10 is a product of 5 and 2. There are lots of 2's in X!. So, you need only worry about the number of 5's in X!.

Thus, 15! will have a 5 each from 5,10,15. Hence the answer is 3.
For 20!, the answer is 4.

Be careful for 25. Here, we are dealing with higher power of 5, i.e., 25 contributes two 5's.
Similarly for 126! the anwer is:
126/5 + 126/25 + 126/25 (consider only the quotient in each fraction; ignore the remainder.)

hope that helps!

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what is the largest integer k such that 15! is divisible by 20 Dec 2007, 20:52
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parsifal

Similarly for 126! the anwer is:
126/5 + 126/25 + 126/25 (consider only the quotient in each fraction; ignore the remainder.)
parsifal
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Its typo here:
126/5 + 126/25 + 126/125 of course
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what is the largest integer k such that 15! is divisible by 21 Dec 2007, 05:45
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what is the largest integer k such that 15! is divisible by 10^k?


what is the largest integer k such that 20! is divisible by 10^k?
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15/10^k, it is basically asking how many different 10's are possible in the numerator.

(15*14*13*12*11*10*9*8*7*6*5*4*3*2)/10^k

clearly, the total number of 10's will be due to 15*14, 10, 5*4

Thus Maximum k=3.

I would follow this approach, because this is faster for me instead of trying to get into some logical or general formalization.
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what is the largest integer k such that 15! is divisible by 21 Dec 2007, 11:19
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bmwhype2
what is the largest integer k such that 15! is divisible by 10^k?


what is the largest integer k such that 20! is divisible by 10^k?

15/10^k, it is basically asking how many different 10's are possible in the numerator.

(15*14*13*12*11*10*9*8*7*6*5*4*3*2)/10^k

clearly, the total number of 10's will be due to 15*14, 10, 5*4Thus Maximum k=3.

I would follow this approach, because this is faster for me instead of trying to get into some logical or general formalization.
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why is it 15 * 14?

the 15 has one 5 in it. that 5 can be multiplied by any 2 from any of the even numbers.
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what is the largest integer k such that 15! is divisible by 21 Dec 2007, 11:39
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parsifal

Similarly for 126! the anwer is:
126/5 + 126/25 + 126/25 (consider only the quotient in each fraction; ignore the remainder.)
parsifal

Its typo here:
126/5 + 126/25 + 126/125 of course
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just to make sure we're on the same page, the answer is 31?
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what is the largest integer k such that 15! is divisible by 21 Dec 2007, 21:41
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bmwhype2
what is the largest integer k such that 15! is divisible by 10^k?


what is the largest integer k such that 20! is divisible by 10^k?

15/10^k, it is basically asking how many different 10's are possible in the numerator.

(15*14*13*12*11*10*9*8*7*6*5*4*3*2)/10^k

clearly, the total number of 10's will be due to 15*14, 10, 5*4Thus Maximum k=3.

I would follow this approach, because this is faster for me instead of trying to get into some logical or general formalization.
why is it 15 * 14?

the 15 has one 5 in it. that 5 can be multiplied by any 2 from any of the even numbers.
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Yeah, whether you multiply that 5 which is in 15 with any other even number of you multiply 15 with 14, it is going to result in the same thing. Just think ....
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what is the largest integer k such that 15! is divisible by 21 Dec 2007, 21:42
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what is the largest integer k such that 15! is divisible by 10^k?


what is the largest integer k such that 20! is divisible by 10^k?
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For 20! it is only one more "zero" which comes with the integer "20", this k=4.
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what is the largest integer k such that 15! is divisible by 22 Dec 2007, 13:05
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what is the largest integer k such that 15! is divisible by 10^k?


what is the largest integer k such that 20! is divisible by 10^k?
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1 .
for the number count all the 2 and 5 in pairs and divide by 2.
Add the result to the number of 0's .

For eg.
15! has 4 ( 2's and 5's ) [ 15 , 12 , 5 , 2 ]
Divide by 2 => 4/2 = 2
Add by number of 0's , we have 1 = 2 + 1 = 3

20! has ( 2 , 5 , 12, 15 ) again 4 2's and 5's
Number of 0 = 2 [ 10 ,20 ]
2 +2 = 4



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