A bag of candy sits on a table. If two kids share all the candy so that each one gets the same number of pieces, there's one candy left over. If three kids share the same candy equally, there are two candies left over. If four kids share the candy equally, there are three candies left over. If five kids share the candy equally, there are four candies left over. If six kids share the candy equally, there are five candies left over.
The smallest number of candies that two kids could share equally with one left over is 3. Two people could also share 5, 7, 9, 11 . . . candies with the same result.
The smallest number of candies that three kids could share with two left over is 5. Three people could also share 8, 11, 14, 17 . . . candies with the same result.
For four kids, the number of candies in the bag could be 7, 11, 15, 19, and so on.
For five kids, the number of candies in the bag could be 9, 14, 19, 24, and so on.
For six kids, the number of candies in the bag could be 11, 17, 23, 29, 35, and so on.
If the puzzle involved just two, three, four, and six kids, the answer would be 11, because these four sequences have that number in common. You need to find a number that all five sequences have in common.
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