BTGmoderatorDC wrote:$$If\ p\ is\ a\ positive\ integer\ and\ p^2\ has\ total\ 17\ positive\ factors,\ then\ find\ the\ number\ of\ positive\ integers\ that\ completely\ divides\ p^3\ but\ does\ not\ completely\ divide\ p?$$
(A) 16
(B) 17
(C) 21
(D) 23
(E) 24
OA A
Source: e-GMAT
Let us first understand how to find out the number of factors of a number, say N.
Say N = a^x*b^y*c^z...; where a, b and c are prime numbers and x, y and z are positive integers
Note that the # of factors includes 1 and N, too.
So, the # of factors of N = (x + 1)*(y + 1)*(z + 1)...
Coming to the question:
We are given that p is a positive integer and # of factors of p^2 is 17.
Say p^2 = a^x*b^y*c^z...; where a, b and c are prime numbers and x, y and z are positive integers
Thus, (x + 1)*(y + 1)*(z + 1) = 17
Since 17 is a prime number (only two factors: 1 and 17) and each of (x + 1), (y + 1) & (z + 1) is greater than 1, we can have p^2 = a^x
Thus, # of factors of p^2 = x + 1 = 17
=> x = 16
Thus, p^2 = a^16
=> p = a^8
Thus, # of factors of p = 8 + 1 = 9;
=> p^3 = a^24
Thus, # of factors of p^3 = 24 + 1 = 25
=> The number of positive integers that completely divides p^3 but does not completely divide p = 25 - 9 = 16.
The correct answer:
A
Hope this helps!
-Jay
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