Postive factors of n excluding n and 1?

gmattesttaker2 wrote:If positive integer n = abcd, where a, b, c, and d are different positive prime
numbers. How many different positive factors does n have, excluding n & 1?

(A) 6
(B) 10
(C) 14
(D) 15
(E) 16

Hey Sri,

Here's a useful rule:

If N = (p^a)(q^b)(r^c)..., where p, q, r,...(etc.) are prime numbers, then the total number of positive divisors of N is equal to (a+1)(b+1)(c+1)...

Example: 14000 = (2^4)(5^3)(7^1)
So, the number of positive divisors of 14000 = (4+1)(3+1)(1+1) = 5x4x2=40

In your question, n = abcd
In other words, n = (a^1)(b^1)(c^1)(d^1)
So, the number of positive divisors of n = (1+1)(1+1)(1+1)(1+1)
= (2)(2)(2)(2)
= 16
So, n has a TOTAL of 16 different positive factors.
The question tells us to EXCLUDE two of the factors (1 and n).

So, the correct answer is 14

Answer: C

Cheers,
Brent

gmattesttaker2 wrote: If positive integer n = abcd, where a, b, c, and d are different positive prime
numbers. How many different positive factors does n have, excluding n & 1?

(A) 6
(B) 10
(C) 14
(D) 15
(E) 16

Another approach is to plug in some numbers.
Let's say that a = 2, b = 3, c = 5 and d = 7 (four DIFFERENT prime numbers)

So, n = (2)(3)(5)(7) = 210
Now let's list ALL of the positive factors of 210: 1, 2, 3, 5, 6, 7, 10, 14, 15, 21, 30, 35, 42, 70, 105, and 210
There is a TOTAL of 16 different positive factors.
The question tells us to EXCLUDE two of the factors (1 and 210).

So, the correct answer is 14

Answer: C

Cheers,
Brent

Hi Sri,

I'm a big fan of TESTing Values on this type of question, but there's also a Number Property concept that could be used here….

Since A, B, C and D are all DIFFERENT PRIME numbers, and N = AxBxCxD, then we know that ALL of the following will divide into N:

1

A
B
C
D

AB
AC
AD
BC
BD
CD

ABC
ABD
ACD
BCD

ABCD

The prompt tells us to exclude 1 and ABCD, so we're left with 14 options.

Final Answer: C

GMAT assassins aren't born, they're made,
Rich