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Two numbers are said to be relatively prime if they share no common

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Two numbers are said to be relatively prime if they share no common [#permalink]

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Two numbers are said to be relatively prime if they share no common positive factors other than 1. Set S contains the set of integers from 1 to 1,000, inclusive. What is the probability that a number chosen from S is relatively prime to 1,000?

A. 5/7
B. 3/5
C. 4/7
D. 2/5
E. 2/7
[Reveal] Spoiler: OA

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Bunuel wrote:
Two numbers are said to be relatively prime if they share no common positive factors other than 1. Set S contains the set of integers from 1 to 1,000, inclusive. What is the probability that a number chosen from S is relatively prime to 1,000?

A. 5/7
B. 3/5
C. 4/7
D. 2/5
E. 2/7


Hi,

Everything depends on the UNDERSTANDING part..

which integers will be CO_PRIME to 1000?
Any integer that does not contain any common factor of 1000 except 1..

what are the PRIME factors in 1000?
2 and 5..

so numbers div by 2 = 1000/2 = 500..
numbers div by 5 = 1000/5 = 200, BUT this 200 includes 100 even multiples which have already been catered for in multiples of 2, So 100..

........(OTHER way = LCM of 2 and 5 =10, so multiples of 10 = 1000/10 = 100
since these 10 have been catered in both 2 and 5 subtract 100 from total 500+200..
)

so TOTAL co-primes =1000-500-100 = 400
prob of picking it = 400/1000 = 2/5
D
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Re: Two numbers are said to be relatively prime if they share no common [#permalink]

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New post 11 Apr 2016, 20:44
we need all numbers between 1 and 1000 that are co-prime.
between 1 to 10 there are 4 : 1,3,7,9
Take the period of 10's , we have 100 periods of 10's between 1 to 1000
So the total number of co-primes = 400
Now, the simple part ...
Probability = 400/1000 (i.e picking a co-prime from the first 1000 numbers )

Ans: 2/5 D

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Re: Two numbers are said to be relatively prime if they share no common [#permalink]

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New post 26 Jun 2017, 20:31
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Lets see what 1000 is made up of (prime factorisation).

1000 = 2^3 * 5^3.. so it is made up of three 2's and three 5's.. more important it is made up of 2 and 5.
This means from 1 to 1000, all those integers which have either 2 or 5 (or both) in their prime factorisation, will NOT be relatively prime with 1000. Or, those integers which are divisible by either 2 or 5 (or both) will NOT be relatively prime with 1000, rest all WILL BE relatively prime with 1000.

So now our task is to find those integers which are NOT relatively prime with 1000, or all those integers (From 1 to 1000) which are divisible by either 2 or 5 or both.

Now, from 1 to 1000, number of integers divisible by 2 = 500... list 1
Also, from 1 to 1000, number of integers divisible by 5 = 200... list 2
And, from 1 to 1000, integers divisible by both 2 and 5 = 100 (these are those which are divisible by 10)

Now, these last 100 integers are present in list 1 as well as list 2. So, in total, all the integers divisible by either 2 or 5 or both =
list 1 + list 2 - 100 (since these 100 are repeated in both lists)
= 500 + 200 - 100 = 600.

So, these 600 are NOT relatively prime with 1000.
Those which ARE relatively prime with 1000 = 1000-600 = 400

So the required probability = (no of favorable outcomes)/(total possible outcomes) = 400/1000 = 2/5

Hence D answer

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Re: Two numbers are said to be relatively prime if they share no common [#permalink]

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New post 13 Nov 2017, 05:14
amanvermagmat wrote:
Lets see what 1000 is made up of (prime factorisation).

1000 = 2^3 * 5^3.. so it is made up of three 2's and three 5's.. more important it is made up of 2 and 5.
This means from 1 to 1000, all those integers which have either 2 or 5 (or both) in their prime factorisation, will NOT be relatively prime with 1000. Or, those integers which are divisible by either 2 or 5 (or both) will NOT be relatively prime with 1000, rest all WILL BE relatively prime with 1000.

So now our task is to find those integers which are NOT relatively prime with 1000, or all those integers (From 1 to 1000) which are divisible by either 2 or 5 or both.

Now, from 1 to 1000, number of integers divisible by 2 = 500... list 1
Also, from 1 to 1000, number of integers divisible by 5 = 200... list 2
And, from 1 to 1000, integers divisible by both 2 and 5 = 100 (these are those which are divisible by 10)

Now, these last 100 integers are present in list 1 as well as list 2. So, in total, all the integers divisible by either 2 or 5 or both =
list 1 + list 2 - 100 (since these 100 are repeated in both lists)
= 500 + 200 - 100 = 600.

So, these 600 are NOT relatively prime with 1000.
Those which ARE relatively prime with 1000 = 1000-600 = 400

So the required probability = (no of favorable outcomes)/(total possible outcomes) = 400/1000 = 2/5

Hence D answer


I'm honestly surprised as to why this reply does not have more "Kudos". It is clearer and more detailed than the other explanations provided for this question.
You have my thanks, my dear individual.

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Re: Two numbers are said to be relatively prime if they share no common   [#permalink] 13 Nov 2017, 05:14
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