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# If each of the digits can be used, as many times as

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If each of the digits can be used, as many times as [#permalink]

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25 Feb 2008, 11:55
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If each of the digits can be used, as many times as necessary, what is the probability of creating a four-digit number that is divisible by four and that begins and ends with a prime number?
(A) 1/18
(B)1/25
(C)2/25
(D) 1/9
(E) 4/25
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25 Feb 2008, 12:26
bmwhype2 wrote:
If each of the digits can be used, as many times as necessary, what is the probability of creating a four-digit number that is divisible by four and that begins and ends with a prime number?
(A) 1/18
(B)1/25
(C)2/25
(D) 1/9
(E) 4/25

= (4 x 10 x 5 x 1)/3000
= 1/15

hmmm. whats the source?
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25 Feb 2008, 12:33

got it off the net.
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25 Feb 2008, 12:34
why are you dividing by 3,000?
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25 Feb 2008, 12:38
1
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GMAT TIGER wrote:
bmwhype2 wrote:
If each of the digits can be used, as many times as necessary, what is the probability of creating a four-digit number that is divisible by four and that begins and ends with a prime number?
(A) 1/18
(B)1/25
(C)2/25
(D) 1/9
(E) 4/25

= (4 x 10 x 5 x 1)/3000
= 1/15

hmmm. whats the source?

1000's place = 2, 3, 5 and 7. so 4 times
100's place = all 10 digits so 10 times
10's place = only 1, 3, 5, 7 and 9. so 5 times
unit place = only 2 so 1 time

so multiply them = 4x10x5x1 = 200.
total = 9x10x10x10 = 9000 (not 3000)

= 200/9000
= 1/45
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25 Feb 2008, 12:41
maratikus wrote:
why are you dividing by 3,000?

i am getting recovered from the lost world.

i do calculation in my mind. for a while i was out of practice so i am practicing it again.
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25 Feb 2008, 12:54
how are u accounting for divisibility by 4?
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25 Feb 2008, 13:19
1
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GMAT TIGER wrote:
1000's place = 2, 3, 5 and 7. so 4 times
100's place = all 10 digits so 10 times
10's place = only 1, 3, 5, 7 and 9. so 5 times
unit place = only 2 so 1 time

so multiply them = 4x10x5x1 = 200.
total = 9x10x10x10 = 9000 (not 3000)

= 200/9000
= 1/45

that's exactly how I solved it too.
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25 Feb 2008, 13:20
bmwhype2 wrote:
how are u accounting for divisibility by 4?

the last digit has to be equal to 2, the third digit has to be an odd number (22 is not divisible by 4, 12 is).

The second digit doesn't matter, the first digit is either 2,3,5 or 7. The second and third digits don't impact divisibility by 4 because 100 is divisible by 4.
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25 Feb 2008, 13:32
i dont understand why the units digit must be the digit 2.

a number is divisible by 4 when the last two digits are divisible by 4
xx04
xx08
xx12
xx16
xx20
etc...
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25 Feb 2008, 14:14
1
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bmwhype2 wrote:
i dont understand why the units digit must be the digit 2.

the number is divisible by 4 -> the last digit has to be even
the last digit is a prime number -> the last digit is equal to 2 -> the only even prime number
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25 Feb 2008, 17:25
good question. we need more of these here. where exactly online did you get it from?
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25 Feb 2008, 21:14
bmwhype2 wrote:
i dont understand why the units digit must be the digit 2.

a number is divisible by 4 when the last two digits are divisible by 4
xx04
xx08
xx12
xx16
xx20
etc...

remember: the question says the last digit can only be a prime. so nothing other than 2 can be in the unit digit.
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25 Feb 2008, 23:04

In this question, we should make 4-digits Number ; hereinafter OOOO

Based on the condition, the last 2-digits of number must be as follows,
if 4-digits numbers can be diviable by 4

OO12
OO32
OO52
OO72
OO92

The begining number must be the prime number, accodring to the condition.

2OOO
3OOO
5OOO
7OOO

the second number can be any number from zero to nine (o~9)

Here is the reasoning for this question.

Probability = (4X10X5) / (9X10X10X4) = 1/18

Is that clear?
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25 Feb 2008, 23:06
GMAT TIGER wrote:
bmwhype2 wrote:
i dont understand why the units digit must be the digit 2.

a number is divisible by 4 when the last two digits are divisible by 4
xx04
xx08
xx12
xx16
xx20
etc...

remember: the question says the last digit can only be a prime. so nothing other than 2 can be in the unit digit.

thanks. this one was tougher than i expected.
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Re: digits   [#permalink] 25 Feb 2008, 23:06
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