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 LCM 540 (Posted on 2014-12-22)
How many pairs of positive integers, without regard to order, have a least common multiple of 540?

 See The Solution Submitted by Charlie No Rating

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 No Subject | Comment 2 of 6 |

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==>220  to get a "match", only the bold digits (belonging to 231 )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
 Posted by Ady TZIDON on 2014-12-22 11:36:32

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