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

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 4862
GPA: 3.82
In the 6-digit integer 543, 2xy, x and y are chosen from the digits 0 [#permalink]

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14 Jan 2018, 23:30
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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 $$8$$?

$$A. \frac{1}{5}$$
$$B. \frac{6}{25}$$
$$C. \frac{7}{25}$$
$$D. \frac{8}{25}$$
$$E. \frac{9}{25}$$
[Reveal] Spoiler: OA

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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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15 Jan 2018, 00:53
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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15 Jan 2018, 10:07
1
KUDOS
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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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 4862
GPA: 3.82
Re: In the 6-digit integer 543, 2xy, x and y are chosen from the digits 0 [#permalink]

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17 Jan 2018, 00:05
=>

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 $$\frac{8}{25}$$

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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, 00:05
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