Some children were seated around a large round table. They pass around a box of candy containing 25 pieces. Bob took the first piece. Each child takes a piece of candy as the box is passed around. Bob also gets the last piece, and he may have more than the first and last piece.
How many children could be seated around the table.
Now, I really don't get this...and I need some help to start....but I think there mightbe 18 children other than Bob....thats just a guess.
There is more than one answer to this.
If there are two people, Bob would get every odd-numbered piece, and 13 in all. If there are 12 children, Bob will get the first, thirteenth and 25th. If there are 8 children, Bob will get the 1st, 9th, 17th and 25th. If there are 24 people, Bob will get the first and 25th piece. If he always gets the last piece, the possible numbers of children (counting Bob) is 2, 3, 4, 6, 8, 12 and 24 -- any number evenly divisible into 24.
To approach this problem, let's break it down step by step.
First, let's assume there are "n" children seated around the table (including Bob), and the box of candy has 25 pieces. Bob takes the first piece, which means there are now 24 pieces left.
As the box is passed around the table, each child takes one piece. Since there are "n" children sitting around the table (including Bob), after one complete rotation, there will be 24 - n pieces left in the box.
Now, we know that Bob also gets the last piece. So, if there are "n" children, how many pieces will be left for Bob to get at the end?
Since each child takes one piece, after one complete rotation, there will be 24 - n pieces left. However, Bob also gets the very last piece, which means there will be (24 - n) - 1 = 23 - n pieces left for Bob.
If Bob has more than the first and last piece, it means he has at least two pieces. Therefore, we can set up an inequality to solve for the possible values of "n". The inequality is:
2 ≤ 23 - n
Let's solve this inequality:
2 + n ≤ 23
n ≤ 21
So, the maximum value of "n" can be 21, which means there can be a maximum of 21 children seated around the table (including Bob). If there are 21 children, there will be 23 - 21 = 2 pieces left for Bob to get, which satisfies the condition that he has more than the first and last piece.
Therefore, your initial guess of there being 18 children (other than Bob) is incorrect. The correct answer is that there can be a maximum of 21 children seated around the table (including Bob).