I have been very remiss about updating my personal website & have not

yet posted the solution to my December brainteaser. Here it is.

The brainteaser read as follows:

Given any four numbers a, b, c, and d, you can form six “pairwise

products”: ab, bc, cd, ad, ac, and bd. I have four particular numbers

in mind. Five of their pairwise products (not necessarily in the order

I just listed) work out to 2, 3, 4, 5, and 6. What does the sixth

pairwise product work out to?

Solution:

If you take the pairwise products in pairs like this:

ab, cd

ac, bd

ad, bc

and multiply together the pair in each row, you of course get abcd

every time. Since only one row contains the unknown pairwise product,

the other two rows, which contain only known pairwise products,

multiply to the same result, namely abcd. Quick inspection shows that

the only possibility is

2, 6

3, 4

giving abcd = 12. The unknown pairwise product must therefore be

12/5, to make the third row come out to 12 also.

(The actual numbers a, b, c, and d are actually irrational –

multiples of the square root of 10. I didn’t ask about that, though.

Concentrate on the problem as stated!)