What is the total number of positive integers that are less

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?

A. 30
B. 40
C. 50
D. 60
E. 70

Bunuel · Accepted Answer

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.

PareshGmat · Answer

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

mau5 · Answer

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.

srinjoy28 · Answer

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...

jlgdr · Answer

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

KarishmaB · Answer

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)

GMATinsight · Answer

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

Fish181 · Answer

how many numbers less than 100 don't have a 2 or 5 as a factor? List out the all the numbers to 50 that work. Once you're half way, you'll notice 20 numbers work. Double that for the "less than 100" goal and you get 40 or b.

1 3 7 9 11 ->5
13 17 19 21 23 -> 10
27 29 31 33 37 -> 15
39 41 43 47 49 -> 20

1 3 7 9 11 ->5
13 17 19 21 23 -> 10
27 29 31 33 37 -> 15
39 41 43 47 49 -> 20

Nadezhda1999 · Answer

There is a very nice formula for this type of questions.

F(n)=n(1-1/p1)*(1-1/p2)...

n is the number in the question 
p1,p2 etc are the prime factors of n

What is the total number of positive integers that are less

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What is the total number of positive integers that are less

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