Catch the fraud!

Question

There are 10 gold smiths in the town and a rich businessman wants to collect 1kg of pure gold. So he informs each person to bring 10 balls of pure gold, each weighing exactly 10g. So altogether he plans to collect 100 gold balls 10g each.

But one gold smith brings 10 balls each weighing exactly 9g.

The businessman gets to know about the ongoing scandal and plans to catch this fraud. He calls the 10 smiths into his office room where he has an electronic balance. The businessman catches the fraud with weighing exactly one time. How did he do that? Note: He weighs one time only. Balls cannot be distinguished by looking.

Well, technically, if all the balls are made of pure gold - and only pure gold - then the $9g$ ball will be smaller... — , May 02 '15 at 10:43
This is a well-known classic: He takes 1,2,3,...,10 balls from the smiths and weights them. Some versions eliminate the weight of the fake balls and only tell you they weight the same. Then one is supposed to determine which are fake and what is the difference in weight. — , May 02 '15 at 10:49
While the businessman watches the scale, if the first goldsmith puts one of his balls on the scale, then the next goldsmith puts one of his balls on the scale.. — , May 02 '15 at 10:51
The fact that you specify it is an electronic balance makes me think one could abuse the "tare" function - and argue that this is only zeroing the balance, not actually making a weighting... Semantics :) — , May 02 '15 at 10:55
@Mattos the scale can only be used 1 time. In your method it have to be used ten times — , May 02 '15 at 10:58
@Demosthene The only reason I used an electronic balance is because it can give you precise reading upto few decimal points. So a small value won't go unnoticed :) — , May 02 '15 at 11:02
@Demosthene The density of the gold is $19.3 g/cm^3$. The fake balls will be $\sqrt[3]{\frac{3\cdot10}{\pi\cdot19.3\cdot4}}-\sqrt[3]{\frac{3\cdot9}{\pi\cdot19.3\cdot4}}=0.0171951...cm$ smaller in radius, a 50th of a millimeter change in radius. People won't notice. — , May 02 '15 at 11:11
@KateGregory yes it is, but wasn't originally posted on P.SE but on Math.SE so he could not have known ;) — Vincent, May 02 '15 at 11:26
I know. Nonetheless it should be closed as a duplicate so that all the answers are in one place. — Kate Gregory, May 02 '15 at 11:28
@VincentAdvocaat Still needs to be closed. Closing a question isn't an insult to the asker (though I know it can feel like one sometimes). — Rand al'Thor, May 02 '15 at 11:29

score 3 · Answer 1 · edited May 02 '15 at 11:29

He takes 1 ball from the first pile, 2 from the 2nd, 3 from the 3th, and so on.

He then weighs them all together, normally he would have a weight of 550 (1 + 2 + 3 + 4 etc times 10g) the difference in weight is the pile it is in.

so if the balls in the first pile were wrong it would weigh 449, if those in the 2nd pile were wrong it would weigh 448 etc.

It's a very well known puzzle and there are a few more similar ones posted on puzzling.stackexchange. If you like these (or other) riddles, check out the site seeing you're new here ;)

Catch the fraud!

1 Answers1