"Extraordinary," said the wizened little shopkeeper. "Such a strange design. What are they?"
"I'm not sure," said Colorado Smith. "They appear to be mythical beasts, of a kind I have never seen before. Five identical golden statues… You will appreciate my inability to say where they originated."
The shopkeeper nodded. The statues were certainly striking, with their forked tongues, double tails and six sets of wings. Not to mention the clawed teeth. His shop was stocked with a great variety of curiosities, and very few of them were of known provenance. He picked up one of the golden statues and weighed it in his hand. "This one is hollow, my friend."
"Yes," said Smith. "They look identical, but their weights vary. Each weighs a whole number of jin, and the total is 13 jin."
"I will need to know their individual weights before I can decide on a price," said the little man. "Unfortunately my weights were stolen last night, and all I have is the balance."
"I can't remember the exact weights," said Smith. "But I did notice something interesting, which you can check using your balance. If you remove any statue from the set of five, then it is possible to divide the remainder into two groups, which balance each other perfectly."
The shopkeeper stroked his skimpy beard. "Ah. In that case, the weights must be—"
What are the five weights?
Scroll down for the answer
The answer
Let the weights be a, b, c, d, e in ascending order. They are all whole numbers, and some may be equal. Their sum is 13. Therefore they cannot all be equal, since 13÷5 is not a whole number. They cannot all be different, since then the sum would be at least 1+2+3+4+5 =15. At least one weight must be 1, because the remaining possibilities can easily be listed and do not work. If a and b are different, then simple calculations show that removing them in turn leads to
conditions that are logically contradictory.
So a=b=1. Removing e, the only possibility is 1+c=1+d, so c=d.
Removing c, either 1+e=1+c (so e=c) or e=2+c.
But if e=c then removing 1 leads to a contradiction, so e=2+c.
Therefore the weights are 1, 1, c, c c+2. Since the sum is 13, c must be 3.
The winner was Giles Cattermole from Cardiff.
