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The hundreds digit of the five-digit number 73

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The hundreds digit of the five-digit number 73 [#permalink] New post 18 Nov 2018, 10:48
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81% (00:40) correct 18% (00:35) wrong based on 11 sessions
The hundreds digit of the five-digit number 73,__95 is missing. If the hundreds digit is to be randomly selected from the ten digits from 0 to 9, what is the probability that the resulting five-digit number will be a multiple of 3?

A. \(0\)

B. \(\frac{1}{10}\)

C. \(\frac{3}{10}\)

D. \(\frac{2}{5}\)

E. \(\frac{1}{2}\)
[Reveal] Spoiler: OA

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Re: The hundreds digit of the five-digit number 73 [#permalink] New post 25 Nov 2018, 11:36
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Explanation:

As per divisibility rule "A number is divisible by 3 if the sum of the digits is divisible by 3"

Here,

73,_95 : the sum of the digits add up to = 24, which is divisible by 3, therefore, one possibility of the blank is 0

Since the sum of the digit adds up to 24 That means by adding 3 will provide the next number a multiple of 3 i.e the blank can be filled up by 0, 3, 6 and 9

Hence the probability that the resulting five-digit number will be a multiple of 3 = \(\frac{4}{10} = \frac{2}{5}\) (Total possibilities is 10 i.e from 0 to 9)
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Re: The hundreds digit of the five-digit number 73   [#permalink] 25 Nov 2018, 11:36
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The hundreds digit of the five-digit number 73

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