A chemistry professor asked his graduate student to find the weights of five different samples of an unknown material.
The student weighed each possible pair of samples, and recorded the weight of each pair as 120, 110, 117, 112, 115, 113, 114, 121, 116, and 118 grams. When he tried to figure out the five individual weights, he realized that he had neglected to record which samples made up each pair.
The professor said that he could figure out the five weights from the data the student had already collected.
How much did the samples weigh?
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Mike Boughton wrote the original puzzle:
> There is an old one called the 12 penny problem: You have 12 coins, all
> apparently alike, but one is either heavier or lighter. Your job is to
> find which and which one in 3 weighings on a balance using only the coins
> themselves.
>
> This version does not require any math. Somehwere I saw a response that
> does. Not only did it contain the solution for 12, which takes a few
> minutes to figure out, but the general solution of how many weighings it
> takes for n coins.
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A pesar de que ambos lados de la balanza tiene dos protones y tres neutrones, los productos de fusión de la derecha han perdido algo de masa. La masa perdida, se pierde con la energía liberada. La cantidad exacta se puede calcular por la famosa ecuación de Albert Einstein, E = mc2.
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