If n is a positive integer, how many different factors n has?

hi ,

can anyone please explain second option here.

Howcome any interget has only two different prime factor.. Is this even correct?

Any number by default will have unity and number itself prime factor.. how "only two prime factor" makes sense here?


Now if we go ahead by assuming that, question is talking about prime factors..

then in that case we'll have four factor..wont we?

1- n/5 = prime
=> n = prime x 5
all of the factor of n can be: 1, prime, 5, and n. So 4. But then if this prime number is 5, then n has only 3 factor: 1, 5, and n.
INSUFF
(also we can test 10 and 25 for this case)

2- N has only two different prime factors.

I understand this bolded phrase as: if n has 2 different prime factor 2 and 5, n can be written as: \(2^x5^y\) where total number of factor of n = (x+1)(y+1)
Again INSUFF, because:
x=1, y=1, n = 10: 4 factors
x=2, y=1, n = 20: 6 factors

1+2:
n = prime x5
n has only 2 different prime factors
Combine together we can be sure the exponent of 5 and the other prime factor is 1, and that prime factor cant be 5.
So n has 4 different factor.
SUFF

\(Statement 1:\) \(\frac{N}{5} = P\), or \(N = 5*P\)
Now \(5\) is prime and if the other the prime no \(P = 5\), then \(N = 5^2\), hence we will have \((2+1) = 3\) factors: 1, 5 & 25
But if the other prime \(P = 3\), then \(N = 5*3\), hence we will have \((1+1)*(1+1) = 4\) factors: 1, 3, 5, 15
Hence \(Insufficient\)

\(Statement 2:\) let's assume two different prime factors are \(2 & 3\). but we are not given what are the exponents of these prime numbers. for e.g
if \(N = 2^1*3^1\), then the number of factors will be \((1+1)*(1+1) = 4\) (i.e 1, 2, 3 & 6), BUT
if \(N = 2^2*3^1\), then the number of factors will be \((2+1)*(1+1)= 6\) (i.e 1, 2, 3, 4, 6 & 12)
Hence \(Insufficient\)

Combining 1 & 2
we know that \(N = 5*P\) and from statement 2 we know that \(N\) has two different prime factors, hence \(P\) is not equal to \(5\).
so the number of factors will be \((1+1)*(1+1) = 4\)
\(Sufficient\)

Option \(C\)

Hi pratik1709

Statement two simply states that N has two different PRIME Factors and NOT two different factors which is very much possible
for e.g \(6 = 2*3\), has only two prime factors \(2 & 3\), but has 4 factors = 1,2,3 & 6

hi

Why it has to be assumed that, 5 and p are not raised to any power in the equation, N = 5 * P, when combining statement 1 and statement 2 together ....?

thanks in advance ....

Hi gmatcracker2017

We know "\(P\)" is prime, hence it cannot be raised to any other power because in that case it will no more be a prime.. for eg if \(P=2\) and you raise it to power \(3\), then \(2^3 = 8\) = not prime

target find factors of n which is a +ve integer

#1
N/5 is a prime number.
N = 10,15,35,25
factors can be 4,3 insufficient
#2
N has only two different prime factors
we can have two different factors but here power values can vary ; insufficient
from 1 &2
one of the prime number will be '5' and other can be any prime number 2,3,7,11
so factor will be 4 always
sufficient
option C

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