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

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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chetan2u · Answer

Divisibility by 2 is if last digit is divisible by 2.
Divisibility by 4 or 2^2 is if last two digits are divisible by 4.
Divisibility by 8 or 2^3 is if last three digits are divisible by 8 and so on.

In general 2^n divides a number if number formed by last n digits of that number is divisible by 2^n.

Here 8 is 2^3, so last three digits 2xy should be divisible by 8.

2xy = 200 + xy, where 200 is divisble by 8, so check for xy only.

Possible values..
00, 08, 24, 40, 48, 64, 80, 88

Total = 5×5 = 25

Prob = 8/25

D

findingmyself · Answer

For a number to be divisible 8, the last three digits has to be either 0 or divisible by 8

For 2xy to be divisible by 8, we have following combinations possible. Given in the question doesnt talk about repetition, we can assume that repetition is allowed.

The total ways in which 2 numbers can be chosen from the set is 5*5=25 ways
Numbers which match the criteria are: 200, 208, 224, 240, 248, 264, 280 and 288

Answer is 8/25

P_Tripathy · Answer

Do we assume here that x and y can be equals? Or like, repetition is allowed? Will GMAT specify in repetition is allowed or not? I didn't count 200 at first thinking the respective values should be different.

findingmyself · Answer

If the question says "without replacement' explicitly in the stem is the only case where repetition is allowed, If the same is not mentioned, then repetition is always allowed.

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

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

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