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?†
To solve this problem, we can work backwards and find the least number of marbles in the bag that satisfies all the given conditions.
Let's start with the last scenario where the marbles are divided among six children and there is one marble left. This means that the total number of marbles must be one more than a multiple of six.
So, we can list down a few numbers that satisfy this condition:
7 (6 marbles divided among 6 children with 1 left)
13 (12 marbles divided among 6 children with 1 left)
19 (18 marbles divided among 6 children with 1 left)
25 (24 marbles divided among 6 children with 1 left)
...
Now, let's consider the second-to-last scenario where the marbles are divided among five children and there is one marble left. This means that the total number of marbles must be one more than a multiple of five.
Now, we need to find the common numbers from the two lists we made earlier (multiples of six with one more marble and multiples of five with one more marble).
If we examine the two lists, we can see that the only number common to both lists is 31 (30 marbles divided among 5 children with 1 left, and 30 marbles divided among 6 children with 1 left).
Therefore, the least number of marbles in the bag is 31.
4*5*6+1 = 121
check:
121 = 4*30+1 = 5*24+1 = 6*20+1