Here is a math problem for you, from the 1971 edition of Polya's How to Solve It, a classic on mathematical problem-solving:
Bob has ten pockets and 44 silver dollars. He wants to put his dollars into his pockets so distributed that each pocket contains a different number of dollars. Can he do so?
1 comment:
I say no -- if you presume that all pockets have coins, then no -- you would need 45 = Sum(i,i=1-10)coins. But, if one pocket's allowed to be empty, then you're still stuck: the minimum the first 9 could have is: 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36, so your last pocket would have to repeat the 8. For each one you increase Pocket #9, the 10th pocket decreases one, repeating a number. Increasing more than one number only makes your problem worse. Do I win a prize?
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