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GRE Math Challenge #15 - If one number is chosen at random

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GRE Math Challenge #15 - If one number is chosen at random [#permalink] New post 09 Sep 2014, 01:39
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If one number is chosen at random from the first 1000 positive integers,then what is the probability that the number is a multiple of 2 and 8?

[Reveal] Spoiler: Answer
1/8

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Re: GRE Math Challenge #15 - If one number is chosen at random [#permalink] New post 26 May 2016, 06:32
Please explain i didn't get the answer?
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Re: GRE Math Challenge #15 - If one number is chosen at random [#permalink] New post 27 May 2016, 05:43
Any number that is a multiple of both 8 and 2 is a multiple of 8. Because of the repetitive pattern of multiple of 8 among numbers, it should not make a difference whether we take 1000 as a sample or 10. There is one multiple of 8 between 1 and 10. so the probability is 1/8
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Re: GRE Math Challenge #15 - If one number is chosen at random [#permalink] New post 24 Aug 2016, 06:28
Expert's post
rapsjade wrote:
Any number that is a multiple of both 8 and 2 is a multiple of 8. Because of the repetitive pattern of multiple of 8 among numbers, it should not make a difference whether we take 1000 as a sample or 10. There is one multiple of 8 between 1 and 10. so the probability is 1/8


Be careful - the part in green isn't true.
If the question were about the first 10 positive integers, then the probability would be 1/10, since only 1 of the first ten positive integers is divisible by 8 (that integer being 8)


The integers from 1 to 1000 inclusive have a nice feature:
1, 2, 3, 4, 5, 6, 7, 8, ___ 9, 10, 11, 12, 13, 14, 15, 16,___ 17, 18, 19, 20, 21, 22, 23, 24, ___ 25, 26, ...

As we can see, in every batch of 8 integers, exactly 1 is divisible by 8.
This patter continues all the way to 1000.
We have: ...___985, 986, 987, 988, 989, 990, 991, 992 ___ 993, 994, 995, 996, 997, 998, 999, 1000
So, all of our batches have exactly 8 integers.
As such, we can conclude that the probability is 1/8 that a selected number is divisible by 8

However, in the numbers from 1 to 10 inclusive, we have: 1, 2, 3, 4, 5, 6, 7, 8, ___ 9, 10
Here each batch does not have exactly 8 integers
So, we cannot conclude that the probability is 1/8 that a selected number is divisible by 8

Cheers,
Brent
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Re: GRE Math Challenge #15 - If one number is chosen at random [#permalink] New post 24 Aug 2016, 06:43
125/1000

The best way to do this is to take all multiples of 8 till 125, as those will directly be divisible by 2 and uve got your answer.
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Re: GRE Math Challenge #15 - If one number is chosen at random [#permalink] New post 13 Sep 2016, 11:34
Hello,

Small doubt,
I had entered the value 0.125 instead of 1/8. Would that be a problem ?

Thanks :)
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Re: GRE Math Challenge #15 - If one number is chosen at random [#permalink] New post 18 Jan 2017, 23:58
the general rule to find number of terms is ((last term - first term)/difference between two consecutive terms) + 1 , so in our example the solution is ;
(1000-8)/8 + 1 = 124 + 1 = 125 terms
probability = 125/1000
Re: GRE Math Challenge #15 - If one number is chosen at random   [#permalink] 18 Jan 2017, 23:58
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