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07 Apr 2016, 08:37
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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
A. 5/7
B. 3/5
C. 4/7
D. 2/5
E. 2/7
26 Jun 2017, 21:31
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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
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
07 Apr 2016, 21:09
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Bunuel
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
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
