You are the ruler of a great empire and you have decided to throw another celebration in 24 hours. One of your most hated rivals will be there, and you decide to use some deadly poison that will definitely kill him in 10 to 20 hours. To reduce suspicion, you decide to also spike your drink with some non-lethal poison that will make you sick in 10 to 20 hours.
Out of 1250 glasses, you insert the deadly poison in one glass and your non-lethal poison in another glass. But because of your forgetfulness, you forgot which glasses had the poisons! Certainly, you don't want another one of your beloved guests to drink a poisoned glass, or even worse, giving yourself the lethally poisoned glass.
Fortunately, you still have your supply of death-row prisoners who won't mind lending a helping hand by taste-testing the glasses. What is the smallest number of prisoners needed to successfully locate both the lethally and non-lethally poisoned glasses?
Note: if the lethal poison acts before the non-lethal poison, then the prisoner will die without any symptoms.
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There are 1250 glasses and we need to find 2 of them. If there were only one to discover, we could order them and translate their number to base 2. So we would need 11 people, each one represents one power if 2. So the fist glass (00000000001) only the person of the power 0 will drink. If there is a 1 on your power house you drink, if there is a 0, you don't. But if we use this logic for 2 glasses, no one could be just sick. For example: glass 00000000011 is poisoned but we do not know witch one is the non lethal one. But if we translate the glasses numbers to base 3, we manage to discover each glass is lethal and the one non lethal poisoned. So the biggest number on base 3 will be 1201022, so we need just 1 person for the power 729 and 6 for the others. To discover both glasses we need at least 13 people. So we need less than 18 to discover the lethal one, independently if we want to know the non lethal one or not .