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

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CEO
Joined: 15 Aug 2003
Posts: 3339
Let n be the number of different 5 digit numbers, divisible  [#permalink]

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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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Intern
Joined: 03 May 2003
Posts: 27
Re: PS : 5 digit numbers  [#permalink]

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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

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

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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

thanks

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

n =24 * 8 =192

thanks
praetorian
SVP
Joined: 30 Oct 2003
Posts: 1681
Location: NewJersey USA

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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
Non-Human User
Joined: 09 Sep 2013
Posts: 11013
Re: Let n be the number of different 5 digit numbers, divisible  [#permalink]

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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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