# Catch the fraud! [duplicate]

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... – Demosthene May 2 '15 at 10:43
• Umm... Asumme they are made out of a less dense alloy. :) – slhulk May 2 '15 at 10:46
• 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. – Alamos May 2 '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.. – Mattos May 2 '15 at 10:51
• I know. Nonetheless it should be closed as a duplicate so that all the answers are in one place. – Kate Gregory May 2 '15 at 11:28

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 ;)