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

9514 1404 393

  (2, 3, 5)  or  (1, 1, n) for n ≥ 1

Step-by-step 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

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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

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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).

answer: a

step-by-step explanation:

10.2 units

-6 is the correct option from the above given options