December 2005
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In my December diary I posed the following brain-teaser.
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?
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Solution
If you take the pairwise products in pairs like this:
ab,
cd
ac,
bd
ad,
bc
then the product of the two numbers on each line is abcd. All three products are therefore equal. However,
only one line contains the
unknown pairwise product. The other two lines, composed entirely of known pairwise products, multiply to the same
result, abcd. The only
possibility with the numbers given is
2,
6
3,
4
giving abcd = 12. The third line must be
5,
x
(x standing for the unknown pairwise product); and this must also multiply to 12. Therefore
x = 12/5.
The numbers a, b, c, d themselves are actually irrational — multiples of
√10. I
didn't ask for them, though!