K=wxyz, where w, x, y, z are prime numbers. Excluding 1 and

Question

K=wxyz, where w, x, y, z are prime numbers. Excluding 1 and itself, how many factors does K have?

A) 6
B) 8
C) 10
D) 12
E) 14

TeHCM · Answer

My answer is A

Just try using 2, 3, 5 and 7

gamjatang · Answer

K=wxyz, where w, x, y, z are prime numbers. Excluding 1 and itself, how many factors does K have? 

4C1 = 4
4C2 = 6
4C3 = 4

4+6+4=14

I pick (E).

remgeo · Answer

ans =14

Taking example 2, 3, 5, 7
Factors are 

2, 2*3, 2*5, 2*7, 2*3*5, 2*3*7,2*5*7
3, 3*5, 3*7, 3*5*7, 
5, 5*7,
7

remgeo · Answer

Can you please explain this.

christoph · Answer

14...

wxyz => w^1*x^1*y^1*z^1 => (1+1)*(1+1)*(1+1)*(1+1) => 16 is the number of factors of K. substract 1 and itself. it gives 14 !

laxieqv · Answer

his solution is like this:
since 1 and the K itself can't be considered factors of K, these factors can't be formed by none of w,x,y,z or can't be formed by w*y*x*z ---> the rest cases are : factors formed by 1 or 2 or 3 of the 4 numbers w,y,x,z.
formed by 1: 4C1
formed by 2: 4C2
formed by 3: 4C3 

in all : 4C1+4C2+4C3

rlevochkin · Answer

no. of factors of w^p*x^q*y^r*z^t=(p+1)(q+1)(r+1) where w,x,y,z are prime factors of K. No. of factors of K=(1+1)*(1+1)*(1+1)*(1+1)=2^4=16 Excluding 16 and 1, No. of factors of k=14. E

K=wxyz, where w, x, y, z are prime numbers. Excluding 1 and

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