The question as it appears in the 02/13 edition of Specs & Techs from GlobalSpec:
You have six coins with the following values (you live in a remote country of course!): 3 cents, 9 cents, 10 cents, 12 cents, 13 cents, and 14 cents. You use five of them to buy two articles. One costs twice as many cents as the other. What is the value of the coin you kept?
(Update 9:15 Am EST 02/20/07) And the Answer is....
If a, b, c, d, and e are the five coins used to buy the articles, then, their sum can be expressed as
a + b + c + d + e = 2x + x = 3x
where x is the cost of the less expensive article. From the above equation we see that the sum of these five coins is a multiple of 3.
Now, the sum of all the coins is: 3 + 9 + 10 + 12 + 13 + 14 = 61 which is 20 * 3 + 1. In other words, 61 is 20 + 1 (mod 3).
Therefore, the coin left in your hands has to have a value of 1(mod 3).
The only two coins in the group that give a value of 1(mod 3) are 10 and 13.
Let's look at the two cases:
(a) Assume that the coin you kept is 13. Then the sum of the other five must be 3x, where x is less expensive product bought. This, however is not possible because 3 + 9 + 10 + 12 + 14 = 48 = 16 * 3, but no combination of coins in this group will give 16.
(b) Now assume the coin left is 10. Then 3 + 9 + 12 + 13 + 14 = 51 = 17 * 3. The combination of coins to buy the products will be: 14 + 3 = 17 = x and 13 + 12 + 9 = 34 = 2x.

Re: Six Coins: Newsletter Challenge (02/13/07)