If N = p1^a + p2^b + p3^c

then ordered factor pairs coprime to each other = (2a + 1)(2b + 1)(2c + 1)

Unordered factor pairs coprime to each other = {[(2a + 1)(2b + 1)(2c + 1) - 1]/2} + 1 (for (1,1)

360 = 2^3 * 3^2 * 5

Ordered factor pairs coprime to each other = (2 * 3 + 1) * (2 * 2 + 1) (2 * 1 + 1) = 7 * 5 * 3 = 105

Un ordered factor pairs coprime to each other = (105 - 1)/2 = 52 + 1 = 53

if we don't consider (1,1) then 52.

Confused!

Detailed approach (Hemanth Venkatesh)

On factorizing 360 we get the following factors:

360 = (2^3) * (3^2) * (5)

First suppose that 5 divides neither co prime factor. Then we have two cases:

Case 1: One factor must be 2 to some positive power while the other factor must be 3 to some positive power. Hence there are 3 * 2 = 6 of these (as there are 3 possible positive powers of 2 and there are 2 possible positive powers of 3).

Case 2: One factor is 1 while the other factor is a divisor of (2^3) * (3^2).

Hence there are (3+1)(2+1)=12 of these.

Now, note that all of these pairs consist of different numbers except the pair {1,1}. For each of the pairs with two distinct elements, we can multiply either element by 5 to get two new co prime pairs of factors (that is the solution {a,b} gives rise to the solutions {5a,b} and {a,5b}.

Finally, the pair {1,1} under this process only gives us one new pair: {1,5}.

In conclusion, we have 3(6 + 12 − 1) + 2 = 53 co prime pairs.