np - prime numbers

Factors of n^ 4 :

n,n^2,n^3,n^4,1 = 5 divisors/factors

Factors of p^3:

p,p^2,p^3,1 = 4 divisors/factors

Factors of np:

n,p,np,1 = 4 divisors/factors



Hence, (E) should be the answer. Can you pls confirm?

If a and b are primes such that a^m and b^n then number of factors of a and b can be expressed as (m+1) and (n+1)

a^m.b^n=(m+1)*(n+1)

n and p are primes and have exactly 4 positive divisors.

This means power should be 3 or 2 terms raise to power of 1.

n^3 has (3+1) or 4 factors.

np= n^1.p^1=2.2=4 factors

Pick E

sairamGmat wrote:If n and p are different positive prime numbers, which of the integers n^4, p^3 and np has (have) exactly 4 positive divisors?
(A) n^4 only
(B) p^3 only
(C) np only
(D) n^4 and np
(E) p^3 and np

The algebraic solutions are great, but almost certainly the quickest way to solve is by picking numbers.

Just let n=2 and p=3:

n^4 = 2^4 = 16... the factors of 16 are 1, 2, 4, 8, 16... not exactly 4, so eliminate (A) and (D).

p^3 = 3^3 = 27... the factors of 27 are 1, 3, 9, 27... exactly 4, so eliminate (C).

np = 2*3 = 6... the factors of 6 are 1, 2, 3 and 6... exactly 4, so choose (E)!

Any prime number has only 2 divisors.

If we square a prime no., number of divisors will always be greater than 2. eg 2 has 2 divisors(1 and 2), but 2^2 has 3 divisors(1,2,4).

Similarly as we increase the power(exponent) divisors will increase. eg 2^5 = 32 have (1,2,4,8,16,32).

Thus when two primes are multiplied it will lead to exactly 4 divisors(1, prime1, prime2, product).

Anyways its always better to key-in numbers.

p^3 has 4 factors

np

say n= 3 and p =2

6 has 1,2,3, 6 (4 factors)

The answer is E.

The rule is:- For distinct prime number a,b,c

Let there is a number k = a^p * b^q * c*r
The number of divisors of number k = (p+1)(q+1)(r+1).


Best Regards,
JOHN

I 'll Mark E.

I 'll Mark E.

option E

Since n and p are positive prime numbers, the product np has 4 factors.
Apart from 1 and itself, n and p are also factors of np.
So total - 1, itself (product np), n and p.

p^3 will have 4 factors - apart from 1 and itself, p and p^2 are factors.

n^4 will have 5 factors - 1, itself, n^3, n^2 and n.

My pick - E

straight E.

p^3 and pn

Ans is (E)..
p^3 has divisors: 1, p, p^2 and p^3
np has divisors: 1, n, p, np

n^4 has 5 divisors

I have marked E.
np is a sure answer for any prime number s it is true.
and p^4 cant be the case as power greater than 3 will not give exactly 4 divisors.
So we are left with E.