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.
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)/(4mn)
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 20161101 13:27:52 