Problem Solving

If n=4p, where p is a prime number greater than 2, how many different positive even divisors does n have, including n?

A) 2
B) 3
C) 4
D) 6
E) 8

Is it safe to plug in? Is there a quick algebraic approach?

Lula,

Solution A

In general, the number of positive factors can be calculated from the prime factorization. If x = m^a * n^b * p^c... where m,n and p are the distinct prime factors of x, then the number of factors of x is (a+1)*(b+1)*(c+1)... (the values of m,n,p... don't matter).

For instance, 45 = 3^2 * 5^1, so it must have (2+1)*(1+1) = 6 positive factors. They are {1, 3, 5, 9, 15, and 45}

In this case, n=2^2 * k^1.

Number of factors: (2+1)*(1+1) = 6 factors.
Odd factors: {1, k} = 2 odd factors.
Even factors = 6-2 = 4 even factors.

The answer is C.


Solution B

Since all values of n must result in the same answer, plugging in an easy value for n is definitely what I would opt for. The answer is C. I go through the question in detail in the full solution below (taken from the GMATFix App).



-Patrick