Gmat_mission wrote: ↑

Tue Nov 10, 2020 7:31 am

For a certain positive integer \(N, N^3\) has exactly \(13\) unique factors. How many unique factors does \(N\) have?

A. 1
B. 2
C. 3
D. 4
E. 5

Answer: E

Solution:

We can analyze the problem by the number of unique primes N has.

If N has only one unique prime factor, let N = p^m where p is a prime. Then N^3 = p^(3m) has 3m + 1 unique factors. Since we are given that N^3 has 13 unique factors, we can create the equation:

3m + 1 = 13

3m = 12

m = 4

In this case, N = p^4 has 4+ 1 = 5 unique factors.

If N has two unique prime factors, let N = p^m x q^n where p and q are two unique primes. Then N^3 = p^(3m) x q^(3n) has (3m + 1)(3n + 1) unique factors. However, since we are given that N^3 has 13 unique factors and if we set (3m + 1)(3n + 1) = 13, we can see that either m or n has to be 0 because 13 = 1 x 13. So N = p^m (if n = 0) or N = q^n (if m = 0). Either way, we see that N has only one unique prime factor, which we’ve discussed above. Furthermore, we can see that N can’t have more than two unique prime factors, either. Therefore, N must be some unique prime raised to the 4th power, and thus it has 5 unique factors.

Answer: E