Four boys sit down to play a game of marbles. They empty the bag of marbles and divide the marbles among themselves, and they find they have one marble left. A fifth boy comes along and wants to play, so they divide the marbles among all five boys and once again find that there is one marble left. A girl comes along and wants to play, so they divide the marbles among all six children and once again find that there is one marble left. What is the least number of marbles that were in the bag?

numbers divisible by 4 with one remainder:

5 9 13 17 21 25 29 33 37 41 45 49 53 57 61 65 ...
numbers divisible by 5 with one remainder:
6 11 16 21 26 31 36 41 46 51 56 61 66 ..
numbers divisible by 6 with one remainder:
7 13 19 25 31 37 43 49 55 61 67 ...

looks like 61 is the smallest number common to all 3 sets

so there were 61 marbles.

To solve this puzzle, let's work backwards by examining the given information. We know that with the first division, there was one marble left. Then, when the fifth boy joined, they divided the marbles among five boys and once again found one marble left. Finally, with the addition of the girl, they divided the marbles among six children and had one marble left.

We can conclude that the number of marbles must be one more than a multiple of 5 (when the fifth boy joined) and one more than a multiple of 6 (when the girl joined). To find the least number of marbles, we need to find the least common multiple (LCM) of 5 and 6, and add one to it.

The LCM of 5 and 6 is 30. Adding one to it gives us the least number of marbles that were in the bag: 30 + 1 = 31. Therefore, the least number of marbles in the bag is 31.