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How many four digit numbers divisible by 4 can be made with

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Director
Joined: 03 Jul 2003
Posts: 652
How many four digit numbers divisible by 4 can be made with [#permalink]

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06 Jan 2004, 17:05
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

How many four digit numbers divisible by 4 can be made with the digits 1, 2, 3, 4, 5 if the repetition of digits is not allowed?
Director
Joined: 14 Oct 2003
Posts: 583
Location: On Vacation at My Crawford, Texas Ranch

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06 Jan 2004, 17:16
Any number (regardless of number of digits) with the last 2 digits that are divisble by 4 is divisble by 4

out of 1,2,3,4,5

xx12, xx24, xx32, xx52

thus 3!x4 = 24
Director
Joined: 03 Jul 2003
Posts: 652

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06 Jan 2004, 17:19

I missed 24. How one can make sure that he/she gets all the numbers
divided by 4 in these kind of questions. These kind of questions tend to
take more time for me. Any suggestions?
Director
Joined: 14 Oct 2003
Posts: 583
Location: On Vacation at My Crawford, Texas Ranch

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06 Jan 2004, 17:31

I missed 24. How one can make sure that he/she gets all the numbers
divided by 4 in these kind of questions. These kind of questions tend to
take more time for me. Any suggestions?

Just know the characteristics of some base numbers - that will save you some time.

e.g., multiples of 2,3,4,5,..etc.

Like I said - multiples of 4 - the last two digits will always be multiples of 4.

Multiples of 2 will always be even (obviously)

Multiples of 5 will always end in 0 or 5.

Multiples of 10 will always end in 0.
Senior Manager
Joined: 30 Aug 2003
Posts: 322
Location: dallas , tx

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08 Jan 2004, 14:30
okay,,this is the problem i was referrign to...
_________________

shubhangi

Director
Joined: 14 Oct 2003
Posts: 583
Location: On Vacation at My Crawford, Texas Ranch

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08 Jan 2004, 14:39
shubhangi wrote:
okay,,this is the problem i was referrign to...

Shubanghi,

Yes it shouldn't be 3! per se just (3x2) - which conveniently is the same as 3!

thus for xx12 3 letters for the first number x 2 letters that we can choose from the numbers that are left.
08 Jan 2004, 14:39
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