August 13th, 2015
Today's Question - A group of seven vagabonds came upon an apple orchard late at night. They agreed that the next morning they would pick all the apples and each would keep what he picked. Two waited for the others to doze off and then set about to pick the apples. When they were all picked, they divided them in half but had one left over. Their bickering awoke a third. They divided by three but there was one left over. A fourth awoke and again equal division had one left over. The same thing happened when shared by five or six – there was always one left over. Only when all seven shared equally was there no remainder. What is the smallest number of apples that could have produced these results – divided by any number from two through six and leaving one leftover?
Reveal Answer
Answer \xe2\x80\x93 To solve the vagabond\xe2\x80\x99s apple problem you could start with two assumptions. First, it had to be an odd number (or it would have easily been divided by the first two). Second, it had to be a multiple of seven since that was the only time there was not \xe2\x80\x9cone\xe2\x80\x9d left over. I first shot to 49 which didn\xe2\x80\x99t quite work. The answer lay much further down the multiplication table. The number is 301.Doesn't match Missing info Like question