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LCM 540 (Posted on 2014-12-22) Difficulty: 3 of 5
How many pairs of positive integers, without regard to order, have a least common multiple of 540?

  Submitted by Charlie    
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Solution: (Hide)
540 = 5 * 2^2 * 3^3.

The 5 can be in either of the integers, or both: that's 3 possibilities.

The 2^2 can be in either or both of the integers, but if it's in only one of them, the other integer could have or not have a factor of 2. So the next factor by which to multiply the possible arrangements is (2*2 + 1), the solitary 1 being the case of both having the 2^2 as a factor, and the 2*2 for the choice of which one has the 2^2, and whether the other integer is or is not even.

So far we have 3*(2*2+1).

The 3^3 can be in either or both the integers. If it's in only one or the other, the one it's not in could be a multiple of 9, or of just 3, or no factor of 3 at all. So the next factor by which we multiply the possibilities is (2*3+1).

That makes 3*(2*2+1)*(2*3+1) = 105.

However, that considers the order of the integers: 1*540 and 540 * 1 are considered separate instances.

The only pair of integers that has not been counted twice is 540*540.

So 105 = 104 + 1, but the 104 must be halved, as we don't want order to matter.

The answer is 52 + 1 = 53.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
re(2): One for column A, and one .. (spoiler?)Steve Herman2014-12-23 10:32:59
Hints/Tipsre: One for column A, and one .. (spoiler?)Ady TZIDON2014-12-22 19:03:11
SolutionOne for column A, and one .. (spoiler?)Steve Herman2014-12-22 16:50:37
re: No SubjectCharlie2014-12-22 12:31:37
No SubjectAdy TZIDON2014-12-22 11:36:32
No SubjectDej Mar2014-12-22 09:50:10
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