If N is a positive integer less than 200, and 14N/60 is an

Let's choose a nice value of N that satisfies the given information.

GIVEN: 14N/60 is an integer
Prime factorize the numerator and denominator to get: (7)(2)(N)/(2)(2)(3)(5) is an integer
Simplify: (7)(N)/(2)(3)(5) is an integer
Notice that, when N = 30 (aka the product of 2, 3, and 5), then (7)(N)/(2)(3)(5) = (7)(2)(3)(5)/(2)(3)(5) = 7, which IS an integer
So, N = 30 satisfies the given information.

N has how many different positive prime factors?
30 = (2)(3)(5)
So, N has 3 different positive prime factors (2, 3 and 5)

Answer: B

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"n" has to be multiple of 60 if 14n/60 has to be an integer. The multiples of 60 < 200 are 60,120,180.

The prime factors of 60 are 2,3,5 . It would be the same with 120 and 180.
The number of prime factors are 3. Hence B.

Pls correct me if my explanation is wrong.

For 14n/60 to be an integer, n only needs to be a multiple of 30.
But the same logic holds even in that case.
As a multiple of 30, it already has 2,3,5 as factors
For 7 to be a factor (next smallest prime), n would need to be 210 which is not possible
Hence the answer is 3.

The lowest prime number, 2, is also the only even prime number. 2, 3, 5, 7, 11…

Every number is made up of at least one prime factor, except for the number 1.

Every number that is not a prime can be broken into prime factors.

Of course this is very basic, but it is important to keep in mind when factoring larger numbers. For instance if we take the number 60 we can see that it is comprised of the factors 4 x 3 x 5. Notice that 3 and 5 are both primes, however 4 is not. A number that is not a prime can always be broken down into more than one prime number, whether those number or numbers are distinct. Therefore 4 can be broken down to 2 and 2.

When taking apart larger numbers sometimes a factoring tree can be helpful. (The U.S. emphasizes this skill and thus it comes naturally for those schooled in the U.S. ). With a large number sometimes the easiest way to approach it is by dividing by 2 if it is even, and if odd, knowing the divisibility rules for 3, 5, etc.

Let’s take a random number, say 136. We can start dividing by 2 as follows: 136/2 = 68, 68/2 = 34, 34/2 = 17. Now we have the prime factors. Three ‘2s’ and a ’17.’ Sometimes a question, such as the question in the thread, will ask for distinct or different primes. In the case of 136, the distinct primes will be 2 and 17.

This of course is really high-level and unless you are at the 200-300 GMAT level you would never such a question. Nonetheless, these fundamentals apply even to difficult prime factorization problems.

So back to the question at hand:

14n/60, can be reduced to 7n/30. Because 7n/30 has to be an integer, n has to be a multiple of 30. The prime factors of 30 are 2, 3, and 5.

The next important part to the question is “different positive factors.” So if we multiply the prime factors 2, 3, 5 times 30, we do not change the number of different prime factors. But as soon as we multiply n times the next highest prime factor, 7, we go over 200: n = 30x7 = 210. Therefore n contains only three prime factors: 2, 3, and 5.

Hope that helped!

14n/60, lets factorize this as 2x7xn / 2x2x3x5 -> 7xn / 2x3x5. Therefore, n has 2, 3, 5 for this to be an integer. That is n needs to be a factor of 30.

3 prime factors, B.

Here first we further break down the no 14n/60 to 7n/30

now 30 = 2*3*5

so to get an integer the numerator must be a multiple of 2*3*5
so the smallest number which satisfies this condition is 30
thus we have 3 prime numbers here for n i.e 2,3,5

so ans is B

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