perplexus.info :: Just Math : A peculiar triplet

A peculiar triplet (Posted on 2016-11-01)

This triplet of positive integers has this peculiarity:
A product of any its two numbers divided by the 3rd number
has 1 as a remainder.

Find it.
Show that no other exist.

See The Solution Submitted by Ady TZIDON

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The triple without complete proof

| Comment 1 of 7

There can be at most one even number. It would be convenient if there were a 2.

So try letting the triple be (2,a,b) with a, b odd

We have

2a=nb+1 and

2b=ma+1 for some integers m, n

Solve for a yields

a=(2+n)/(4-mn)

there are only a few pairs of (m,n) to check

m n a b

3 1 3 2

1 3 5 3

1 2 2 1.5

2 1 1.5 2

1 1 1 1

only the second fits so the triple is (2,5,3)

The assumption of a 2 was arbitrary. It doesn't rule out another even or three odds.

Posted by Jer on 2016-11-01 13:27:52

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