VJesus12 wrote: ↑

Sun Jul 12, 2020 12:04 am

If \(p\) is a positive integer and \(p^2\) has total \(17\) positive factors, then find the number of positive integers that completely divides \(p^3\) but does not completely divide \(p?\)

(A) 16
(B) 17
(C) 21
(D) 23
(E) 24

[spoiler]OA=A[/spoiler]

Source: e-GMAT

Note that for a number N = a^n*b^m, the no. of positive factors of N = (n + 1)(m + 1)

Since 17 is a prime number, it can only be written as (16 + 1). Thus, p^2 = a^16; where a is a prime number

=> p = a^8 => No. of factors of p = (8 + 1) = 9;

=> p^3 = a^24 => No. of factors of p^3 = (24 + 1) = 25

Thus, the number of positive integers that completely divides \(p^3\) but does not completely divide \(p\) = 25 – 9 = 16

Correct answer: A

Hope this helps!

-Jay
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VJesus12 wrote: ↑

Sun Jul 12, 2020 12:04 am

If \(p\) is a positive integer and \(p^2\) has total \(17\) positive factors, then find the number of positive integers that completely divides \(p^3\) but does not completely divide \(p?\)

(A) 16
(B) 17
(C) 21
(D) 23
(E) 24

[spoiler]OA=A[/spoiler]

Solution:

We can let p = 2^8 (notice that p^2 = 2^16 has 16 + 1 = 17 factors). In this case, we see that p^3 = 2^24 and if q = 2^n where n is an integer between 9 and 24, inclusive, q divides p^3 but not p. Since there are 24 - 9 + 1 = 16 values for n, there are 16 values for q.

Answer: A