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Re: If k is the number of zeros at the end of n!, where n is a positive in [#permalink]
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srach wrote:
chetan2u wrote:
Bunuel wrote:
If k is the number of zeros at the end of n!, where n is a positive integer less than or equal to 100, then how many values can k take ?

A. 20
B. 21
C. 22
D. 24
E. 25





n! means product of all integers from 1 till n, so 0 will get added whenever a multiple of 5 becomes n.
How many times does n take value of a multiple of 5 => \(\frac{100}{20} = 20\)

But, we also have to add one omre to it, when k is 0, which will be the case from 1! to 4!.

Total = 20+1 = 21

B


Can chetan2u or Bunuel please explain this statement? "But, we also have to add one omre to it, when k is 0, which will be the case from 1! to 4!." I am unable to comprehend it. Thanks in advance.


Added the details. Hope it helps.
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Re: If k is the number of zeros at the end of n!, where n is a positive in [#permalink]
chetan2u wrote:
Bunuel wrote:
If k is the number of zeros at the end of n!, where n is a positive integer less than or equal to 100, then how many values can k take ?

A. 20
B. 21
C. 22
D. 24
E. 25





n! means product of all integers from 1 till n, so 0 will get added whenever a multiple of 5 becomes n.
How many times does n take value of a multiple of 5 => \(\frac{100}{20} = 20\)
These are {xy0, xy00, xy000, xy0000, xy00000, ……….}

But, we also have to add one more to it, when k is 0, which will be the case from 1! to 4!.
Since we are not given that k is a positive integer, it may be possible that k is 0, that is the factorial is just xy
1!=1, 2!=2, 3!=6 and 4!=24…..None of them have a zero in the end, so k is 0.

The moment we increase n to 5, 5!=120…..Here k is 1.

Total = 20+1 = 21

B


chetan2u Bunuel
Can you please explain why you do not consider all the power of 5? I mean in 100! there are 20 power of 5 but there are also 4 power of 25 so a total of 5^24
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Re: If k is the number of zeros at the end of n!, where n is a positive in [#permalink]
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ibnking wrote:
chetan2u wrote:
Bunuel wrote:
If k is the number of zeros at the end of n!, where n is a positive integer less than or equal to 100, then how many values can k take ?

A. 20
B. 21
C. 22
D. 24
E. 25





n! means product of all integers from 1 till n, so 0 will get added whenever a multiple of 5 becomes n.
How many times does n take value of a multiple of 5 => \(\frac{100}{20} = 20\)
These are {xy0, xy00, xy000, xy0000, xy00000, ……….}

But, we also have to add one more to it, when k is 0, which will be the case from 1! to 4!.
Since we are not given that k is a positive integer, it may be possible that k is 0, that is the factorial is just xy
1!=1, 2!=2, 3!=6 and 4!=24…..None of them have a zero in the end, so k is 0.

The moment we increase n to 5, 5!=120…..Here k is 1.

Total = 20+1 = 21

B


chetan2u Bunuel
Can you please explain why you do not consider all the power of 5? I mean in 100! there are 20 power of 5 but there are also 4 power of 25 so a total of 5^24


We are looking at the different possibilities of zeroes in end. There could be 1(5! to 9!), or 2(10! to 14!) and so on.
We are not looking at the maximum number of zeroes.

100! has 24 zeroes as you have mentioned but 99! will have 22, so there is no n! that contains 23 zeroes and similarly there are 3 more whenever we add multiples of 25.
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Re: If k is the number of zeros at the end of n!, where n is a positive in [#permalink]
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Bunuel wrote:
If k is the number of zeros at the end of n!, where n is a positive integer less than or equal to 100, then how many values can k take ?

A. 20
B. 21
C. 22
D. 24
E. 25



This is the approach in which I understand this question:

To get each trailing 0, we need one 2 and one 5 in the mix of numbers in the factorial. So n! can be anything between 1! to 100!. I think to see how many different values of k are possible, we need to see how many 5s can possibly be there.

So, anywhere between
1! to 4!, you would have no trailing zeroes, so k = 0
5! to 9!, you would have one trailing zero, so k = 1
10! to 14!, you would have two trailing zeroes, so k = 2
15! to 19!, you would have three trailing zeroes, so k = 3
20! to 24!, you would have four trailing zeroes, so k = 4
25! to 29!, you would have SIX trailing zeroes (because 25 has TWO fives, not one), so k = 6
and so on until 100!

I think where most people are making a mistake is considering k = 5 which is not possible as seen from above. Similarly, k = 11, 17 and 23 are also not possible since 49! would have ten 5s in it (and hence 10 trailing zeroes) but 50! would have twelve 5s in it (and hence 12 trailing zeroes). Same for 17 and 23.

If we follow this pattern, we will see then that 21 values of k are possible, i.e. all numbers from 0 to 24 (25 numbers) except for 5, 11, 17 and 23 (4 numbers)
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Re: If k is the number of zeros at the end of n!, where n is a positive in [#permalink]
Thank you.
By writing all the values of K I found 21. I misunderstood the questions with the maximum value of k.
In fact K can't take some value 5, 11 ....

1. If n<5 => n! has NO "0"
2. If n=5 => n! has 1"0" so k=1
3. If n=10 => n! has 2 "0" so k=2
4. If n=15 => n! has 3 "0" so k=3
5. If n=20=> n! has 4 "0" so k=4
6. If n=25 => n! has 6 "0" so k=6
7. If n=30 => n! has 7 "0" so k=7
8. If n=35 => n! has 8 "0" so k=8
9. If n=40 => n! has 9 "0" so k=9
10. If n=45 => n! has 10 "0" so k=10
11. If n=50 => n! has 12 "0" so k=12
12. If n=55 => n! has 13 "0" so k=13
13. If n=60 => n! has 14 "0" so k=14
14. If n=65 => n! has 15 "0" so k=15
15. If n=70 => n! has 16 "0" so k=16
16. If n=75 => n! has 18 "0" so k=18
17. If n=80 => n! has 19 "0" so k=19
18. If n=85 => n! has 20 "0" so k=20
19. If n=90 => n! has 21 "0" so k=21
20. If n=95 => n! has 22 "0" so k=22
21. If n=100 => n! has 24 "0" so k=24
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Re: If k is the number of zeros at the end of n!, where n is a positive in [#permalink]
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