Arithmetic - Factors

shankar.ashwin wrote:If 'N' is the minimum number which has 20 different positive integer factors, the value of 'N' is :

A) 240
B) 480
C) 1024
D) 15552
E) 2^21

There's a nice rule that says:
If N = (p^a)(q^b)(r^c)... where p, q, r etc are prime numbers, then the total number of different positive divisors of N = (a+1)(b+1)(c+1)...

Now let's check the answer choices, beginning with the smallest (minimum) value.

240 = (2^4)(3^1)(5^1)
So, the number of different positive divisors of 240 = (4+1)(1+1)(1+1) = 20

So, the answer must be A

Cheers,
Brent

shankar.ashwin wrote:If 'N' is the minimum number which has 20 different positive integer factors, the value of 'N' is :

20 = 2*10 = 2*2*5 = 4*5

We know that if N = (p^a)*(q^b)*(r^c)*... (where p, q, r etc are prime numbers)
Then, number of different positive factors of N = (a + 1)(b + 1)(c + 1)...

Hence, considering 20 = 2*10
N must be of the form p*(q^9).
In this case minimum possible value of N is 3*(2^9) = 3*512 = 1536

Now, considering 20 = 2*2*5
N must be of the form p*q*(r^4).
In this case minimum possible value of N is 3*5*(2^4) = 3*5*16 = 240
This is the minimum of the options. Hence we got our answer.

The correct answer is A.

If 240 was not given in the answer, then we have to check for 20 = 4*5.
In that case N must be of the form (p^3)*(q^4).
In this case minimum possible value of N is (3^3)*(2^4) = 27*16 = 432 > 240

vaibhavgupta wrote: I went for B

B = 480

Since 480 = (2^5)(3^1)(5^1), the number of different positive divisors of 480 = (5+1)(1+1)(1+1) = 24

Cheers,
Brent