How many pairs of positive integers, without regard to order, have a least common multiple of 540?

I am not 100% sure that my result is correct, but I would like to present my process of thought.

Since 540 = 2^2 *3^3 *5^1 there are 24 possible factors , each representable by a triplet (abc) the numbers in the brackets representing the powers of 2,3,5.- from(000) to (231) commas omitted

Now for a factor n, represented by (abc) the matching factors needed to produce LCM=540 are all numbers of a form (ABC) such that

max (a,A=2) and max (b,B)=3 and max (c,C)=1.

Then we count the # of matching candidates for each of the 24 from the lowest to the highest and erase the few repetitions .

The 24 factors are** 1,2,3,4,5,6,9,10,12,15 , 18,20,27,30,36,45,60,90,108, 135, 180, 270, 540**.

*The corresponding counts are ***1,1,1, 3,2 1,1,2, 3, 2 , 1,6,3,2,3,2,3,3,2,4, 2,2,1,1. -duplicity eliminated.**

* 17+29+6=52 pairs (edited)*

**I hope the result is ok;**

I just want to show how the list was compiled, for **36**, say:

36==>**2**20 to get a "match", only the **bold** digits (belonging to **2**31 )may be lower or equal in the "partner", which needs to contribute ***31 ****so we get 031 131 231**

i.e. 3 numbers,(15th in the list).

**As to the correctness of my final number I prefer "waiting and comparing" over "rechecking and debugging".**

**After recounting and reading Steve's solution: **__I believe still 1 pair is missing : the answer should be 53 -__I got 52 and do not have the time to find where is the miss.

*Edited on ***December 22, 2014, 8:44 pm**