Show that for every integer n > 0, there exists a positive integer with precisely n positive divisors.

2^(n-1) has n divisors, specifically 1, 2, 4, 8, ... 2^(n-1)

let us take 2 which is > 0.

so, n = 2

we need to look for a positive integer with exactly 2 divisors [i.e factors]

so take 3 or 5 or any prime.

all of them have exactly 2 positive divisors.

another example .

take 4

so n = 4

there are again many nos. such as 8 which has 1,2,4,8 as its divisors.

if n = 6

there is 12 with 1,2,3,4,6,12 divisors .