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Difficulty:
95%
(hard)
Question Stats:
41% (02:29) correct
59%
(02:43)
wrong
based on 123
sessions
History
Date
Time
Result
Not Attempted Yet
A natural number n is such that 120 ≤ n ≤ 240. If HCF of n and 240 is 1, how many values of n are possible?
A. 24
B. 28
C. 32
D. 40
E. 48
A. 24
B. 28
C. 32
D. 40
E. 48
nick1816
\(120 \leq n \leq 240\)
\(240 = 2^4 * 3 * 5\)
Since HCF of 240 and n is 1, it means n has none of 2, 3 and 5. 120 has 2, 3 and 5 as factors. 120 to 239 are 120 integers.
Of these 120 integers, half will have 2 as a factor so ignore those. We are left with 60 integers.
1/3 will have 3 as factor which means 40 will have 3 as factor. Of these, half will be even and half odd. We have already removed the even integers. So we remove another 20 integers only. Now we are left with 60 - 20 = 40 integers.
1/5 will have 5 as factor which means 24 integers. Of these, half will be even so they have already been removed. We are left with 12 integers. Of these, a third will have 3 as factor so they have already been removed too. We are left with 8 integers.
40 - 8 = 32 integers will have no 2, no 3 and no 5.
Answer (C)
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Joined: 20 Jul 2017
Last visit: 16 Sep 2026
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GMAT 1: 690 Q51 V30
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nick1816
HCF of n and 240 is 1 ---> There should be no common factors between 240 and n
240 = \(2^4\)*3*5
So, n should not have any multiples of 2, 3 or 5
Lets calculate the number of multiples of 2,3 & 5 and subtract from total
Number of multiplies of 2 between 120 & 240 = 61
Number of multiplies of 3 between 120 & 240 = 41
Number of multiplies of 5 between 120 & 240 = 25
Multiples common to 2&3, 3&5, 2&5 twice. The same needs to be subtracted
Number of multiplies common to 2&3 i.e 6, between 120 & 240 = 21
Number of multiplies common to 3&5 i.e 15, between 120 & 240 = 9
Number of multiplies common to 2&5 i.e 10, between 120 & 240 = 13
We need to add the multiple of 2&3&5 once as the subtraction of cases 2&3, 3&5, 2&5 from 2, 3 & 5 removes all the cases of 2&3&5.
Number of multiplies common to 2&3&5 i.e 30, between 120 & 240 = 5
Number of multiples of 2, 3 & 5 between 120 & 240, inclusive = 61 + 41 + 25 - 21 - 9 - 13 + 5 = 89
Non multiples = 121 - 89 = 32!!
Hit Kudos if you like my solution
General Discussion
240= 2^4*3*5, Therefore n must not be multiple of 2,3 or 5
Number of multiples of 2 from 120 to 240 (both inclusive)=[{240-120}/2]+1=61
Number of multiples of 3 from 120 to 240 (both inclusive)=[{240-120}/]3+1=41
Number of multiples of 5 from 120 to 240 (both inclusive)=[{240-120}/5]+1=25
Number of multiples of 6 from 120 to 240 (both inclusive)=[{240-120}/6]+1=21
Number of multiples of 10 from 120 to 240 (both inclusive)=[{240-120}/10]+1=13
Number of multiples of 15 from 120 to 240 (both inclusive)=[{240-120}/15]+1=9
Number of multiples of 30 from 120 to 240 (both inclusive)=[{240-120}/30]+1=5
Total number of multiples of 2,3 or 5 from 120 to 240= 61+41+25-21-13-9+5=89
Possible values of n=121-89=32
Number of multiples of 2 from 120 to 240 (both inclusive)=[{240-120}/2]+1=61
Number of multiples of 3 from 120 to 240 (both inclusive)=[{240-120}/]3+1=41
Number of multiples of 5 from 120 to 240 (both inclusive)=[{240-120}/5]+1=25
Number of multiples of 6 from 120 to 240 (both inclusive)=[{240-120}/6]+1=21
Number of multiples of 10 from 120 to 240 (both inclusive)=[{240-120}/10]+1=13
Number of multiples of 15 from 120 to 240 (both inclusive)=[{240-120}/15]+1=9
Number of multiples of 30 from 120 to 240 (both inclusive)=[{240-120}/30]+1=5
Total number of multiples of 2,3 or 5 from 120 to 240= 61+41+25-21-13-9+5=89
Possible values of n=121-89=32
nick1816
Posts: 4,849
This is an excellent question, which looks like it’s easy, but it’s actually tricky and can be time consuming if you are not familiar with concepts like factors and co-prime numbers.
240 is a common multiple of 2, 3 and 5. So, if n has to be co-prime to 240, it cannot contain any of the above prime factors or their powers. The best way, therefore, is to count the number of multiples of 2, 3 and 5 between 120 and 240 and eliminate them. Whatever is left, has to be the answer.
There are a total of 120 positive integers between 120 and 240. Of these, half are even and half are odd. We do not want the even integers since they will not be co-prime with 240. We are left with 60 odd numbers.
Of these 60 odd numbers, we can have multiples of 3 and 5. We need to eliminate these as well.
Between 120 and 240, we can have 20 even multiples of 3. These have already been removed, so, we need to remove 20 odd multiples of 3 from 60. That leaves us with 40 numbers.
We have 24 multiples of 5 between 120 and 240, of which half are odd, so 12 of them. We have 8 multiples of 15 (3*5) between 120 and 240, of which half are odd, so 4 of them. But these have already been removed when we removed odd multiples of 3. So, we have (12-4) multiples of 5 which cannot be co-prime with 240.
Therefore, (40 – 8) = 32 numbers which are co-prime with 240. The correct answer option is C.
Although the above approach looks a little cumbersome and time consuming (it definitely is time consuming), it is the better approach because it’s collectively exhaustive. Of course, you also have to be good with your knowledge of multiples and factors.
On an average, a question like this can take anywhere between 2 minutes to 3 minutes. So, if you have solved this question within these bounds, it’s a good thing.
Hope this helps!
240 is a common multiple of 2, 3 and 5. So, if n has to be co-prime to 240, it cannot contain any of the above prime factors or their powers. The best way, therefore, is to count the number of multiples of 2, 3 and 5 between 120 and 240 and eliminate them. Whatever is left, has to be the answer.
There are a total of 120 positive integers between 120 and 240. Of these, half are even and half are odd. We do not want the even integers since they will not be co-prime with 240. We are left with 60 odd numbers.
Of these 60 odd numbers, we can have multiples of 3 and 5. We need to eliminate these as well.
Between 120 and 240, we can have 20 even multiples of 3. These have already been removed, so, we need to remove 20 odd multiples of 3 from 60. That leaves us with 40 numbers.
We have 24 multiples of 5 between 120 and 240, of which half are odd, so 12 of them. We have 8 multiples of 15 (3*5) between 120 and 240, of which half are odd, so 4 of them. But these have already been removed when we removed odd multiples of 3. So, we have (12-4) multiples of 5 which cannot be co-prime with 240.
Therefore, (40 – 8) = 32 numbers which are co-prime with 240. The correct answer option is C.
Although the above approach looks a little cumbersome and time consuming (it definitely is time consuming), it is the better approach because it’s collectively exhaustive. Of course, you also have to be good with your knowledge of multiples and factors.
On an average, a question like this can take anywhere between 2 minutes to 3 minutes. So, if you have solved this question within these bounds, it’s a good thing.
Hope this helps!
CEO2021
CEO2021
120 ≤ n ≤ 240 is a total of 240 - 120 + 1 = 121 integers.
But note that these 121 integers include 240 which has HCF of 240 with 240. So we are ignoring it and looking at only the other 120 integers (from 120 to 239).
We can make complete groups of 2 integers, 3 integers and 5 integers if we have 120 integers. It leads to no confusion hence it is better to use 120.
It doesn't matter whether the answer choices are close together or far apart. We will get only one answer and that is 32.
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CEO2021
Adding on to what Karishma has written.
If the extremes are also included, add ONE to the difference of the numbers.
Here both 120 and 240 are to be included, so 240-120+1=121
If only one of the extreme is included, the difference of the numbers will give the quantity.
Say only 240 is to be included, so 240-120=120
So here we have 121 numbers starting with even number, so you will have 61 even numbers and 60 odd numbers.
In the end, the answer will remain the same.
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