Thread: Not sure what to do to solve this?

Supersaver mom Sally came home late one evening with a bagful of DVD’s she bought for $1 each at a yard sale. She planned to split them equally between her three teenage children: Gabe, Jax, and Amanda. It was a school night and past the kids’ bedtime so Sally said that she would divide the DVD’s evenly between her kids in the morning and she left the bag of DVDs on the kitchen table

In the middle of the night, Gabe wanted to make sure he got the DVDs he wanted. Some of the DVDs were action movies; he took a third of them and one extra one.

An hour later, Jax had the same idea so he went to the bag of DVDs. Jax liked comedies and several of the DVDs remaining were what he wanted to see. He took a third of them and one extra one and stored them in his room.

Before sunrise, Amanda woke and quietly tiptoed to the bag of DVDs. She liked romance; she took one third of what DVDs remained and one extra one.

After breakfast, Sally picked up the bag of DVDs to split between her children. She was puzzled why the bag was so light. She was shocked when she opened the bag to find a mere portion of the original DVDs she had brought home for her children.
She divided the remaining DVDs between her children but kept one for herself.


How many DVDs had Sally originally brought home to divide between her children?


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I thought you might be able to use a system of linear equations, but I don't know. I keep thinking that well if the person person was g = 1/3B+1 and then the second person would take j = 1/9B+1, A = 1/27B+1...then the mom gives 1/3 to her kids and keeps one for herself. But the kids are actually drawing from the bag 2/3B-1 instead, I guess and then etc.

Originally Posted by A Beautiful Mind

Supersaver mom Sally came home late one evening with a bagful of DVD’s she bought for $1 each at a yard sale. She planned to split them equally between her three teenage children: Gabe, Jax, and Amanda. It was a school night and past the kids’ bedtime so Sally said that she would divide the DVD’s evenly between her kids in the morning and she left the bag of DVDs on the kitchen table

In the middle of the night, Gabe wanted to make sure he got the DVDs he wanted. Some of the DVDs were action movies; he took a third of them and one extra one.

An hour later, Jax had the same idea so he went to the bag of DVDs. Jax liked comedies and several of the DVDs remaining were what he wanted to see. He took a third of them and one extra one and stored them in his room.

Before sunrise, Amanda woke and quietly tiptoed to the bag of DVDs. She liked romance; she took one third of what DVDs remained and one extra one.

After breakfast, Sally picked up the bag of DVDs to split between her children. She was puzzled why the bag was so light. She was shocked when she opened the bag to find a mere portion of the original DVDs she had brought home for her children.
She divided the remaining DVDs between her children but kept one for herself.


How many DVDs had Sally originally brought home to divide between her children?


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I thought you might be able to use a system of linear equations, but I don't know. I keep thinking that well if the person person was g = 1/3B+1 and then the second person would take j = 1/9B+1, A = 1/27B+1...then the mom gives 1/3 to her kids and keeps one for herself.

There will never be a single solution to a problem like this. In fact, there will be infinitely many possible solutions. Presumably what you are looking for is the smallest possible solution.

If there are DVDs to start with, the first person takes , leaving The second person takes a third of this new number, plus 1 (not knowing that someone else had been there first, and therefore not knowing the original value of x), and leaves two-thirds of them minus 1, namely

The third person similarly leaves two-thirds of the number that she found, less 1, namely The mother then finds that number and divides them up giving to each of the three kids, plus 1 for herself. Therefore Clear the fractions to get You want the smallest solution to this equation, with x and y both positive integers. So the right-hand side of the equation must be a multiple of 8. That should help you work out what must be the smallest admissible value of