K=wxyz
Assuming w,x,y and z are the digits of k then
w,x,y and z can be a value from [2,3,5,7]

So the number of factors will be varying from number to number as we can form 96 different numbers using the 4 digits(Repetition allowed).

Can u please review the question?

16 must the be the answer.

lokesh r wrote:16 must the be the answer.

How? Please explain in detail!!

shovan85 wrote:

lokesh r wrote:16 must the be the answer.

How? Please explain in detail!!

Since nothing is mentioned in the question about the power to which w, x, y, z are raised, i assume it is 1.
So using the formula

say N=w^1 x x^1 x y^1 x z^1

No of factors of N is = (1+1)x(1+1)x(1+1)x(1+1) = 2x2x2x2=16

Not getting it... I was actually predicting that factors are 12 as it is a 4-digit number and max can be 7777 (as 7 is the biggest prime number out of first 10 digits) and 2^12 gives 4096 but 2^13 gives 8192 which is larger than 7777.

But I am not sure I am right.
Any expert with a detailed nice explanation??

K = wxyz, where w, x, y, z are prime numbers. Not including 1 and K, how many factors does K have?

Answer is 22

w, x, y, and z these 4 are factors of K. Also 6 more, wx,wy,wz,xy,xz,yz. And wxy,wxz,wyz (3) each for x, y, and z also which equals 12.
= 4+6+12 = 22

pzazz12 wrote:K = wxyz, where w, x, y, z are prime numbers. Not including 1 and K, how many factors does K have?


I don't know the options for this questions ....

help me to solve this problem

i only count 14 factor without including 1 and k
k=xyzw so k can be divisible by w,y,x,z,wx,wy,wz,xy,xz,zy,wxy,wxz,xyz

dunno how you guy find 16 and 22 and cant understand why....

my pick is 14..... exclude 1 & K .... 16 is the total number of factors including 1 and K

yes. 14 is correct. I repeated some of the combinations.

Yea, it is 14 indeed. Can we solve it like this, 2^4=16 factors - 2 (1 & K)= 14 factors instead of counting all the factors?