Can you find three distinct positive integers A, B and C such that the sum of their reciprocals equals 1?

In other words: 1/A + 1/B + 1/C = 1 where A does not equal B does not equal C (and A does not equal C).

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How about 2, 3, and 6?

ReplyDelete1/2 + 1/3 + 1/6 =

3/6 + 2/6 + 1/6 =

6/6 =

1

You got it right. The interesting part is this is the only combination that works. If they weren't distinct, you could use 1/4 + 1/4 + 1/2 or 1/3 + 1/3 + 1/3. But I believe that's it for possibilities.

ReplyDeleteIf it has to be three of them, then their average value must be 1/3. Without 2 this is impossible unless all three are 3's.

ReplyDeleteSo let's consider:

1/2 + 1/x + 1/y = 1

==> 1/x + 1/y = 1/2

==> (x + y)/xy = 1/2

==> 2x + 2y = xy

==> 2x - xy = -2y

==> x(2 - y) = -2y

==> x = -2y/(2 - y).

==> to x = 2y / (y - 2)

==> x = 2 + 4/(y - 2) [the result of long division, quite useful]

Now, for x to be an integer, y - 2 must divide 4.

So y - 2 = 1, and y = 3 (and x = 6) or

y - 2 = 2, and y = 4 (and x = 4) or

y - 2 = 4, and y = 6 (and x = 3) (repeat of previous solution.

So 3,3,3 or 2,3,6 or 2,4,4 and that's it.

If we opened it up to negatives, we could use -2, 1, and 2

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