0
$\begingroup$

This question already has an answer here:

A king sent 10 of his servants out to buy him each 10 gold rings. Each ring should weigh 10 grams, and altogether there should be 100 gold rings.

One of the King's spies told him that one servant had cheated and had brought 10 rings each of 9 grams instead of the desired 10 grams. Unfortunately, the spy hadn't been able to recognize the face of the servant, so that the king was left wondering which of the servants had stolen his money.

In order to discover the unfaithful servant, the king used a digital balance but only once.

How could he do that?

$\endgroup$

marked as duplicate by Rob Watts, wbogacz, Bozman, mdc32, Gilles Nov 4 '14 at 22:46

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ A King can only use one digital balance only one time? $\endgroup$ – jyoonPro Nov 5 '14 at 8:23
6
$\begingroup$

Spoiler so others can have fun:

Assign each servant a number 1 through 10. Have each servant put that number of rings on the scale. So #1 will put one ring, #2 will put two rings, ..., and #10 will put all 10 rings.
The weight will be off by the number of grams matching the cheating servant.

If nobody cheated the total should be:
(1 * 10g) + (2 * 10g) + (3 * 10g) + ... + (10 * 10g) = 550g

However lets say #2 cheated, so now it's:
(1 * 10g) + (2 * 9g) + (3 * 10g) + ... + (10 * 10g) = 548g

550 - 548 = 2, the number of our cheater!

$\endgroup$
1
$\begingroup$

Take 1 ring from the first servant, 2 rings from the second servant and so on. Put all those rings in a pile. Weigh the pile. If all rings were the same weight, the pile should weigh exactly 550 grams. Since one of the servants brought rings that weigh one gram less, the pile weighs now 550 - x grams. Therefore the culprit is the xth servant.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.