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**Beth****Guest**

Grace has 16 jellybeans in her pocket. She has 8 red, 4 green, and 4 blue ones. What is the minimum number of jellybeans she must take out of her pocket to ensure that she has one of each color?

Thanks.

Beth

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

You should probably start off with some combination of numbers. Then again, just making a whole bunch of puns about it most likely won't help.

Have you heard of the pigeon hole principle? This is kind of like the reverse of that.

If you haven't, think of it this way: Come up with a scheme of pulling jelly beans out (not randomly) so that you try to avoid having all three colors for as long as possible.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,585

Nine jellybeans, oh beth, for 4 plus 4 is eight and one last one is the red one.

**igloo** **myrtilles** **fourmis**

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

If she only picked nine, she could have 8 red beans and one of another colour, meaning she only had two kinds. She'd been to pick 13 to definitely get one of each.

Why did the vector cross the road?

It wanted to be normal.

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,585

Oh, right 8-), yeah add the 8 and the 4 plus 1 more.

**igloo** **myrtilles** **fourmis**

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**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

yeah mathsyperson is very thoughtful, I used to judge it should be all.

**X'(y-Xβ)=0**

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