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If each of the digits can be used, as many times as [#permalink]
 25 Feb 2008, 12:55
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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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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can you explain your answer. got it off the net.
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Director
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why are you dividing by 3,000?
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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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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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how are u accounting for divisibility by 4?
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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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Director
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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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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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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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good question. we need more of these here. where exactly online did you get it from?
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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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 Answer is A) 1/18 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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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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