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The hundreds digit of the five-digit number 73 [#permalink]
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Question Stats: 76% (00:48) correct 24% (00:36) wrong based on 25 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

_________________ Director Joined: 20 Apr 2016
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Re: The hundreds digit of the five-digit number 73 [#permalink]
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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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Word Problem [#permalink]
The hundreds digit of the five-digit number 73,_95 is missing. If the hundreds digit is to be randomly selected from the ten different digits from 0 to 9, what is the probability that the resulting five-digit number will be a multiple of 3 ?
A 0

B 110

C 310

D 25

E 12 Word Problem   [#permalink] 26 Mar 2019, 07:53
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