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Let n be the number of different 5 digit numbers, divisible

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CEO
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Let n be the number of different 5 digit numbers, divisible  [#permalink]

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New post 11 Oct 2003, 05:55
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Let n be the number of different 5 digit numbers, divisible by 4 with the digits 1, 2, 3, 4, 5 and 6, no digit being repeated in the numbers. What is the value of n?

1) 144
2) 168
3) 192
4) None of these

Thanks
Praetorian

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Joined: 03 May 2003
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Re: PS : 5 digit numbers  [#permalink]

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New post 11 Oct 2003, 13:48
praetorian123 wrote:
Let n be the number of different 5 digit numbers, divisible by 4 with the digits 1, 2, 3, 4, 5 and 6, no digit being repeated in the numbers. What is the value of n?

1) 144
2) 168
3) 192
4) None of these

Thanks
Praetorian


To be div by 4 the last 2 digits must be div by four so we have:
_ _ _ _ 12
_ _ _ _ 16
_ _ _ _ 24
_ _ _ _ 32
_ _ _ _ 36
_ _ _ _ 52
_ _ _ _ 64

for each of these we could have 4*3*2*1 (ie. 4*3*2*1 12)
so n = 24 * 7 = 168

please check my work
thanks
CEO
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Joined: 15 Aug 2003
Posts: 3339
Re: PS : 5 digit numbers  [#permalink]

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New post 13 Oct 2003, 01:24
exy18 wrote:
praetorian123 wrote:
Let n be the number of different 5 digit numbers, divisible by 4 with the digits 1, 2, 3, 4, 5 and 6, no digit being repeated in the numbers. What is the value of n?

1) 144
2) 168
3) 192
4) None of these

Thanks
Praetorian


To be div by 4 the last 2 digits must be div by four so we have:
_ _ _ _ 12
_ _ _ _ 16
_ _ _ _ 24
_ _ _ _ 32
_ _ _ _ 36
_ _ _ _ 52
_ _ _ _ 64

for each of these we could have 4*3*2*1 (ie. 4*3*2*1 12)
so n = 24 * 7 = 168

please check my work
thanks


Just missed ...you forgot to do the same for _ _ _ _ 56

n =24 * 8 =192

3) 192 is the answer

thanks
praetorian
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Joined: 30 Oct 2003
Posts: 1681
Location: NewJersey USA
  [#permalink]

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New post 01 Jan 2004, 21:26
1
I believe they should be 6 digit numbers from 6 six digits. six digit number can be expressed as

10^5 * x1 + 10^4 * x2 + 10^3 * x3 + 10^2 * x4 + 10 ^ 1 * x5 + x6

All the powers except 10 * x5 + x6 are divisible by 4, so we need to find
the combination of x5 and x6 amoung 1,2,3,4,5,6 that is divisible by 4
(1,2) (1,6) (2,4) (3,2) (3,6) (5,2) (5,6 ) (6,4) for each of these we can have 4! ways of arraging the remaining numbers
total numbers = 4! * 8 = 192
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Re: Let n be the number of different 5 digit numbers, divisible  [#permalink]

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New post 09 Jul 2017, 03:51
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Re: Let n be the number of different 5 digit numbers, divisible   [#permalink] 09 Jul 2017, 03:51
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