GMAT Club World Cup 2022 (DAY 9): If k is a positive integer and k has

Answer: D

If k is a positive integer and k has 25 distinct positive factors, how many distinct positive factors does 35k have ?

25 = 5*5 = 1*25 = 25*1
If k = a^m * b^n
then, (m+1)(n+1) = 5*5 or 1*25 or 25*1
=> m,n = 4 or m=0, n=24 or m=24, n=0

To calculate distinct positive factors of 35k i.e. 5*7*(a^m)*(b^n)
we need to know if a or b will be any of the value 5 or 7 or not.

(1) Both 5k and 7k have 50 distinct positive factors.
5k = 5*(a^m)*(b^n)
If a = 5, then factors = (m+2)(n+1) = 50
From the given condition, if m,n =4, then (m+2)(n+1) = 6*5 = 30 !=50.. m,n = 4 is not possible.
if m = 0, n = 24, then (m+2)(n+1) = 2*25 = 50 ... possible.
if m =24, n = 0, then (m+2)(n+1) = 26*1 != 50 ... not possible.
Similarly, if b = 5, then factors = (m+1)(n+2)
m=0, n=24 is the possibility.
Now, if we consider a or b = 7, then again we will get m=0, n=24 and m=24, n=0
Since, both 5k and 7k have 50 distinct positive factors,
a and b will be 5 and 7 ... a and b are interchangeable.. it won't make a difference.
i.e. k = (5^m)*(7^n)
Factors for 5k = 50 = (m+2)(n+1) => m=24, n=0
and factors for 7k = 50 = (m+1)(n+2) => m=0, n=24
This is not possible at the same time.
=> k do not have a factor as 5 or 7 i.e. a or b cannot be equal to 5 or 7
Then factor of 5k = 2*factors of k = 2*25 =50
and factor of 7k = 2*factors of k = 2*25 = 50
Therefore factors of 35k
= factors of 5*7* a^m * b^n
= (1+1)*(1+1)*(m+1)*(n+1)
= 2*2*25
=100
Sufficient.

(2) k is a multiple of 363
let k = 363 * a^m * b^n
i.e. k = 3*11^2 *a^m * b^n
Factors of k = (1+1)*(2+1)*(m+1)*(n+1)
= 6*(m+1)*(n+1)
We already know, factors of k = 25 and 25 is not a multiple of 6
=> either of a or b will not be 5 or 7.
So, going back to 35k, k does not have 5 or 7 as a factor.
Therefore, factors of 35k = factors of 5*7*a^m * b^n
= (1+1)*(1+1)*(m+1)*(n+1)
= 4*(m+1)*(n+1)
= 4*25
= 100
Sufficient.