If n is a positive integer, does n have four or more distinct factors?

Thank you, chatan2u! I guess I didn't really know what distinct factors mean. So, now as I understand they are the ones that we get from the factor tree, that is only primes?

(1) n is not prime.
If n is 1, we have only 1 factor. If n is a perfect square we would have 3 distinct factors. If n is a perfect cube we would have 4 distinct factors and so on. Clearly, (1) is INSUFFICIENT.

(2) 150 <= n < 200
150 = 2x3x5^2. Thus, it has > 4 distinct factors
However, 151 and 199 are prime nos. so they have only 2 distinct factors and 169 and 196 are perfect squares so they have 3 distinct factors.
Thus, (2) INSUFFICIENT.

Together (1) & (2) remain insufficient as can be seen from explanations above. Therefore, answer is E.

MAGOOSH OFFICIAL SOLUTION:

Keep in mind, a prime number is number with only two factors, 1 and itself. All positive numbers with only two positive factors are prime numbers.

If we multiple one prime by another prime, say 2*5 = 10, then we get four factors, {1, 2, 5, 10}. Any product of two different primes has four factors. If one of the numbers we multiply is not prime, then it already has more than two factors of its own, and so the resultant product will have many more than 4 factors. It would seem that non-prime numbers have to have at least 4 factors.

The one exception is: squares of prime numbers. Consider what happens when we square 5: we get 25, and the factors of 25 are {1, 5, 25}. The square of a prime number is not prime, and it has three factors.

Statement #1: n is not prime

Well, if n is not prime, n could be 24 or 25. The first, 24, has factors {1, 2, 3, 4, 6, 8, 12, 24}—eight factors, so the answer to the prompt question is “yes.” Meanwhile, 25 has three factors, so the answer to the prompt question is “no.” Two different answers to the prompt are possible. This statement, alone and by itself, is not sufficient.

Statement #2: 150 ≤ n < 200

Clearly, many of the numbers in that range will have more than 4 factors, and the answer to the prompt is “yes.” But here, n could be prime, and there must be some prime numbers in that range. You don’t need to know this, but the prime numbers in this range are {151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199}. If n were any of these numbers, n would have only two factors, and the answer to the prompt question would be “no.” Two different answers to the prompt are possible. This statement, alone and by itself, is not sufficient.

Combined statements:

Now, n must be in the 150 to 200 range, and all prime numbers are excluded. As note, most of the numbers in this range will have a large number of factors, so for almost all of them, the answer to the prompt question is “yes.” The trouble is: there one number in this range that is the square of a prime number. The square of 13 is 169, so the factors of 169 are {1, 13, 169}. Since this is the square of a prime, it has exactly three factors. This gives a “no” answer to the prompt question. Even with combined statements, we still can get two different answers. Even combined, the statements are not sufficient.

Answer = (E)

This explanation was incredibly useful especially to apply to future questions, thanks a lot!

Hi Viktorija,

"Distinct factors" simply means different factors. The distinct factors of 12 are 1, 2, 3, 4, 6, and 12. This term is most useful when we are talking about primes, though, because that's where we are likely to repeat. (No one would ever say "Oh, the factors of 12 are 1, 2, 3, 4, 6, 12, 6, 3, 4, 2, 6, . . . . ") The prime factorization of 12 is 2*2*3, but the distinct prime factors are just 2 and 3--we don't list 2 twice.

Tricky problem--it's really just testing whether we know this particular rule about squares of primes vs. other numbers. However, at least we are pretty likely to get down to C & E, and there's no shame in guessing on a problem like this.

Tried with basic work. taking numbers between 151 - 155.
151 : prime.
152: 2*2*2*19
153: 3*3*17
155: 5* 31

So we got all combinations, 155 not prime, but distant factors are less than 4. 152, 153 are not prime but factors equal to 4 or more

Careful! The distinct factors of a number include 1 and the number itself, so 155 has four factors. As others indicated above, the only non-prime in that range with fewer than 4 factors is 169. You pretty much have to recognize the rules being tested to get this right with confidence.

(Hey look! I just used "being" in a sentence. What would the GMAT do with that?)

Q : n has 4 or more distinct factors?
what we should recall about distinct factors :
> Set 1 :: primes : only 2 distinct factors
> Set 2 :: perfect squares of primes : only 3 distinct factors
> Set 3 :: all others : have 4 or more distinct factors

Thus we need to find the no's which fall under such categories.
1) n is not prime. consider 4 (3 distinct factors + not prime) and 6(4 distinct factors + not prime)
Thus NS.
2) 150<= n <200
in this range we have 169 ie 13*13, which has only 3 distinct factors. Thus NS.

1 + 2 together also do not give any information about excluding Set 2 (eg 169) from range. Thus NS.
Ans E.

Hi Dmitry,

Oops thanks for correcting me. Still lots more to look into

Another High Quality Question.
Here is what i did in this one--->

We need to see if the number of factors of n is greater than or equal to 4 or not.

A few key links =>

1 is the only number that has one factor.

A prime number has exactly two factors.

For any number to have 3 factors it must be of the form => Prime^2 i.e square of a prime number.

Lets get on with the statements=>

Statement 1->
n is not prime.
Lets play around with some test cases.
n=1 => One factor => n has less than 4 factors
n=200=> 2^3*5^2 => 12 factors => n has more than 4 factors.
Hence insufficient.

Statement 2->
n lies in the range [150,200]
Again lets use some test cases here.
n=151=> Prime number as it is not divisible by any prime number less than equal to the square root of 151.
So n has 2 factors which is less than 4.
n=200=> 12 factors => more than 4 factors.
Not sufficient.

Combing the two statements =>
n is not prime.
ans n≠1
So n can never has one or two factors.

But what about three?
Can n be square of a prime ?
YES for n=13^2=169=> n has 3 factors which is less than 4
For all values of n => n will have more than 3 factors.
Hence not sufficient

Hence E.


Similar Question to practise ->
if-n-is-a-positive-integer-does-n-have-exactly-three-factors-195997.html#p1515021

Bunuel why have we not considered negative factors?????

Question: does n have the number of factors >=4?

(1) Not sufficient. take 1 or 4 factors < 4 but for 6 or above factors > = 4

(2) Not Sufficient. If n is a 151, which is prime it will have only 2 factors. If n is 180 then the number of factors is >=4.
for 169 which is not prime factor <= 4

(1)+(2) Wven after prime number not there it is insufficient

Answer E

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