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what is the least N such that N! is divisible by 1000? a)

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what is the least N such that N! is divisible by 1000? a) [#permalink] New post 30 May 2007, 10:43
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C
D
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what is the least N such that N! is divisible by 1000?

a) 8
b)10
c) 15
d) 20
e)25

explanation please.... OA will follow
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 [#permalink] New post 30 May 2007, 10:48
Factor (1000) = 2,5,2,5

we need N! to include at least all numbers 2,5,2*5=10, so N! has to be equal to or grater then 10!

the answer is (B)

:-D
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 [#permalink] New post 30 May 2007, 11:06
prime factors (1000)=5*5*5*2*2*2

N! to include min. of three 5s and a min. of three 2s for becming divisible by 1000.

min. (N!) = 15! ==15......*10......*8........5.....
=5.........5*2.......2*2*2........5 (suff.)
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 [#permalink] New post 30 May 2007, 11:45
OA is C

any other explanations?
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 [#permalink] New post 30 May 2007, 11:48
vidyasagar wrote:
prime factors (1000)=5*5*5*2*2*2

N! to include min. of three 5s and a min. of three 2s for becming divisible by 1000.

min. (N!) = 15! ==15......*10......*8........5.....
=5.........5*2.......2*2*2........5 (suff.)


Yes ! my mistake , I factored 1000 as 2,2,5,5

well done vidyasagar

-----------------------------

another way to explain is:

Factor (1000) = 2,2,2,5,5,5

10! = 2*3*4*5*6*7*8*9*10

2*3*4*5*6*7*8*9*10/2*2*2*5*5*5 = we can cancel out to get:

(3*6*7*8*9*2)/5 = not an integer

15! = 2*3*4*5*6*7*8*9*10*11*12*13*14*15

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

3*6*7*8*9*2*11*12*13*14*3 = an integer

:)
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 [#permalink] New post 30 May 2007, 12:06
Clearly (C)

When you prime factor 1000 you get Three 5's... 15 is the least number that will give you Three 5's.
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Ans C [#permalink] New post 30 May 2007, 13:37
Prime Factor for 1000:
5*5*5*2*2*2

This goes to the Denominator.
So in the numerator we need to have all of this to arrive at the least possible number. That would be 15.

I.e.:
Numerator:
2*3*(2*2)*5*(3*2)*7*(2*2*2)*(3*3)*(5*2)*11*(3*2*2)*13*(7*2)*(3*5)

This is nothing but trying out numbers till I get to the one where all of the denominator gets reduced to 1.

Denom:
5*5*5*2*2*2

So the least is 15. You dont have to expand all of the numbers though.
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Re: Divisible PS... [#permalink] New post 30 May 2007, 20:06
hsk wrote:
what is the least N such that N! is divisible by 1000?

a) 8
b)10
c) 15
d) 20
e)25

explanation please.... OA will follow


I don't know how to explain this one! I just tried a few values and 15 seems to be the answer. Is there some short cut or trick which can be used?
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Re: Divisible PS... [#permalink] New post 30 May 2007, 20:29
LM wrote:
hsk wrote:
what is the least N such that N! is divisible by 1000?

a) 8
b)10
c) 15
d) 20
e)25

explanation please.... OA will follow


I don't know how to explain this one! I just tried a few values and 15 seems to be the answer. Is there some short cut or trick which can be used?


just think about the least value you need , if you know that Factor (1000) is 2,2,2,5,5,5, on the face of it N! has to include at least 5?

since 5! = (5*4*3*2*1) and (5*4*3*2*1)/2*2*2*5 = 3

but you still have 5,5, left !! so using the same logic we can say that N! has to be at least 10! = 10*9*8*7*6*5*4*3*2*1

since the next number that can be canceled out by 5 is 10 , and that leave us with the last five , that can only be canceld out with 15

so the least N! is 15!

:-D
Re: Divisible PS...   [#permalink] 30 May 2007, 20:29
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