Since 100=2^2*5^2 then a number not to have a positive factor in common with 100 other than 1 should not have 2 and/or 5 as a factors.

# of multiples of 2 in the range (98-2)/2+1=49 (check this: totally-basic-94862.html#p730075);
# of multiples of 5 in the range (95-5)/5+1=19;
# of multiples of both 2 and 5, so multiples of 10, in the range (90-10)/10+1=9 (to get the overlap of above two sets);

Hence there are total of 49+19-9=59 numbers which are multiples of 2 or 5;

Total positive integers less than 100 is 99, so there are 99-59=40 numbers which have no positive factor in common with 100 other than 1.

Answer: B.

Hope it's clear.
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The approach taken by you is correct. I took a slightly long approach.
I took the numbers as:
3,5,7,9,11..All the odd numbers starting with 3 till 99. Such numbers total to 49
Then, I took all the numbers divisible by 5. These numbers will have their factor common with 100. The count of such numbers is 19.
In between the above two sets, we have few numbers in common - 5, 15, 25 ..10 in total.
Now, I am confused here. We have the following:
Set 1 - 49
Set 2 - 19
Set 3 - 10
Total numbers = 49-19 = 30
How do we deal with Set 3? We should add it to the above figure, but don't know the exact reasons..where is the overlapping of data that should cause us to add it to the figure of 30. Please help.

Not clear what are you doing here.

There are 50 odd numbers from 1 to 100, not 49.

Next, why are you subtracting from that the number of multiples of 5?
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Basically the question asks about the total no of co-prime factors of 100. Bunuel has already explained the method, however, for just knowing something new, there is another method to do this :

100 = Find out all the prime factors = 2 and 5. Thus total no of co-prime integers to 100, and less than 100 = (1-1/2)(1-1/5)*100 = 1/2*4/5*100 = 40.

So, if I have to find out the total no of co-prime factors for 48, that would be -->

Total prime factors of 48 = 2,3. Thus the co=prime factors less than 48 = (1-1/2)(1-1/3)*48 = 1/2*2/3*48 = 16. This includes 1, which is co-prime to 48.

This is not some thumb rule, there is a proper derivation for this.Though, it is beyond the scope of GMAT.
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Total number of odd integers is 50 of which 10 integers are divisible by 5.

So the correct answer is (B) 40

Formula : Number of integers less N and are co-prime to N is given by : N(1-1/a)(1-1/b)(1-1/c).....where a, b, c are prime factors of N..
In the given equation, the prime factors of 100 are 2 and 5. Hence the number will be 100(1-1/2)(1-1/5) = 40.
Hope it helps...

I did it like this:

There are 50 odd numbers
There are 10 multiples of 5 among those 50 odd numbers

Therefore 50-10 = 40

Answer is B

Could someone confirm if this method is OK

Cheers
J

Another way to solve this problem:
Since 100=2^2*5^2 then an integer to not have a positive factor in common with 100 other than 1 should not have 2 and/or 5 as a factors
Positive integers that do not have 2 and/or 5 as factors would be (all odd numbers MINUS odd numbers that end in 5)
# of odds < 100 = 50
# of odd numbers that end in 5 (5, 15, 25, 35, 45, 55, 65, 75, 85, and 95) = 10

Ans = 50 - 10 = 40

My approach:
All even integers and 100 have 2 as a common factor. Similarly all multiples of 5 and 100 have 5 as a common factor. Rest all the integers and 100 are co-primes.
In between 1 and 10 there are 4 such integers (1, 3, 7, 9), which have no common factor with 100 except 100 and 1.
And for all numbers less than 100, we have 10 such sets 1-10, 11-20, 21-30 .... 91-100

So total number of integers less than 100 that have no common factors with 100 = 4 * 10 = 40
Option B.
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\(100 = 2^2 * 5^2\)

We require to find out numbers which do not have 2 & 5 as factors

Total odd numbers from 1 to 99 = 50 (These do not have 2 as factors)

However the above 50 numbers have 5, 15.... upto 95 (Total =10) factors of 5

Removing the same

50-10 = 40

Answer = B

Question: What is the total number of positive integers that are less than 100 and that have no positive factor in common with 100 other than 1?

My way: First I eliminated 1 and 100, then I eliminated 2,4,5,10,20,25,50 since they're all factors of 100.
Now I don't know how to go next. Since you have to eliminate all the factors of 100, I don't know if I need to eliminate all 2's, all 4's, all 5's and so on...

Please help...

Merging similar topics. Please refer to the discussion above.

P.S. Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Pay attention to rules 1, 2, 3, 7, and 8. Thank you.

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Another way to think about it:

Let's consider numbers from 1 to 100 (since it eases the calculations).

Co-prime with 100 means that they should have no factor of 2 and/or 5.

In the first 100 positive integers, 50 are divisible by 2 (including 100).
So we remove these 50 and are left with 50 numbers not divisible by 2.

Next, in the first 100 numbers, 100/5 = 20 are divisible by 5.
Out of these 20, 10 are even so we have already removed them. We need to remove another 10 with are odd multiples of 5.

We are left with 50 - 10 = 40.

Answer (B)
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For those familiar with Euler's phi function:
phi(100) = phi(2 squared) * phi(5 squared) = (2) * (25 - 5) = 40

100 = 2^2 * 5^2

i.e the Solutions must not contain any factor of 2 and 5 in order to satisfy the given constraints in the question.

Number of multiples of 2 from 1 to 100 = 100/2 = 50

Number of multiples of 5 from 1 to 100 = 100/5 = 20

Number of multiples of 10 from 1 to 100 = 100/10 = 10

So Total number that are multiple of either of 2 or 5 = 50+20-10 = 60

Remaining number = 100 - 60 = 40

Answer: Option B
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The question says: Other than 1...in that sense, its 39. Can anyone clarify?