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In the 6-digit integer 543, 2xy, x and y are chosen from the digits 0

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In the 6-digit integer 543, 2xy, x and y are chosen from the digits 0  [#permalink]

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New post 15 Jan 2018, 00:30
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[GMAT math practice question]

In the 6-digit integer \(543, 2xy, x\) and \(y\) are chosen from the digits \(0, 2, 4, 6\) and \(8\). What is the probability that \(543, 2xy\) is divisible by \(8\)?

\(A. \frac{1}{5}\)
\(B. \frac{6}{25}\)
\(C. \frac{7}{25}\)
\(D. \frac{8}{25}\)
\(E. \frac{9}{25}\)

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In the 6-digit integer 543, 2xy, x and y are chosen from the digits 0  [#permalink]

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New post 15 Jan 2018, 01:53
MathRevolution wrote:
[GMAT math practice question]

In the 6-digit integer \(543, 2xy, x\) and \(y\) are chosen from the digits \(0, 2, 4, 6\) and \(8\). What is the probability that \(543, 2xy\) is divisible by \(8\)?

\(A. \frac{1}{5}\)
\(B. \frac{6}{25}\)
\(C. \frac{7}{25}\)
\(D. \frac{8}{25}\)
\(E. \frac{9}{25}\)



In order to solve the question, we must know the rule for divisibility of 8.
The rule is as follows: If the last 3 digits are divisible by 8, the number is divisible by 8.

Since the last 3 digits are 2xy we need to find out how many of the combinations are divisible by 8.

They are 200,208,224,240,248,264,280,288(8 options)
The denominator must be the total options possible: 5*5 = 25

Therefore, the probability that \(543, 2xy\) is divisible by \(8\) is \(\frac{8}{25}\)(Option D)
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Re: In the 6-digit integer 543, 2xy, x and y are chosen from the digits 0  [#permalink]

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New post 15 Jan 2018, 11:07
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MathRevolution wrote:
[GMAT math practice question]

In the 6-digit integer \(543, 2xy, x\) and \(y\) are chosen from the digits \(0, 2, 4, 6\) and \(8\). What is the probability that \(543, 2xy\) is divisible by \(8\)?

\(A. \frac{1}{5}\)
\(B. \frac{6}{25}\)
\(C. \frac{7}{25}\)
\(D. \frac{8}{25}\)
\(E. \frac{9}{25}\)


Total possible number - xy -> 5X5 = 25

In given question, we have 2xy as 200 is divisible by 8, we just need to worry about xy. (2xy = 200 + xy)

So combination with 0,2,4,6,8 divisible by 8 -> 00,08,24,40,48,64,80,88
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Re: In the 6-digit integer 543, 2xy, x and y are chosen from the digits 0  [#permalink]

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New post 17 Jan 2018, 01:05
=>

An integer with three or more digits is divisible by 8 if and only if its last three digits from a number that is divisible by \(8\).

The values of \(2xy\) that are divisible by \(8\) are \(200, 208, 224, 240, 248, 264, 280\), and \(288\).
The total number of 6-digit integers of the form \(543,2xy\) is equal to the number of ways of choosing two digits from \(0, 2, 4, 6\) and \(8\), which is \(5*5 = 25.\)
Thus, the probability that \(543,2xy\)is divisible by \(8\) is \(\frac{8}{25}\)

Therefore, the answer is D.
Answer: D
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Re: In the 6-digit integer 543, 2xy, x and y are chosen from the digits 0   [#permalink] 17 Jan 2018, 01:05
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