Question

Find three positive whole numbers such that for any two of them, the

number one less than their product is divisible by the third number.

I have no idea what this means

Answer 1

9514 1404 393

Answer:

(2, 3, 5) or (1, 1, n) for n ≥ 1

Explanation:

For some a, b, c, you want to show that ...

(ab -1)/c is a whole number

(ac -1)/b is a whole number

(bc -1)/a is a whole number

__

A search of triples with numbers in the range of 1 to 100 gave these results:

(1, 1, n) . . . for any n ≥ 1

(2, 3, 5) . . . . apparently, the only non-trivial solution

__

Further comment

We can show that if you postulate the triple is (2, x, 2x-1), then the only non-trivial value for x is 3. This gives (2, 3, 5) as above.

This triple requires ...

(2(2x -1) -1)/x = integer = (4x -3)/x = 4 -3/x

In order for 3/x to be an integer, we must have x=1 or 3. For x=1, the triple is (2, 1, 1). For x=3, the triple is (2, 3, 5).