Probability That a Sum Is Divisible by ThreeDate: 11/04/2009 at 07:53:02 From: Sepha Subject: Probabilities divisible by three In how many ways can you choose three numbers from 1-100 whose sum is divisible by three? I don't get how to even start doing this. I tried to do it manually, like finding all the numbers divisible by three, like 3, 6, 9 and 12, etc. and seeing what three numbers can add up to them. Date: 11/04/2009 at 09:19:38 From: Doctor Ian Subject: Re: Probabilities divisible by three Hi Sepha, I'd probably start thinking about ways that I could just look at three numbers, and see if the sum will be divisible by 3. For example, as you've noted, if you choose all three numbers that are individually divisible by 3, the sum will also be divisible by 3. What if you choose two numbers that are divisible by 3, and a third that is not? Try some examples, and convince yourself that this won't work. What about two numbers that aren't divisible by 3? Well, this CAN work, but only if one is 1 less than a multiple of 3, and the other is 1 more than a multiple of 3. And there is a fourth possibility--that none of the numbers is divisible by 3. What can happen then? I'll leave that for you to think about. Now, suppose we have 100 balls, marked with the numbers from 1 to 100. We color all the multiples of 3 green: 3, 6, 9, 12, ... 99. The balls whose numbers are 1 MORE than a multiple of 3 are colored blue: 1, 4, 7, 10, 13, ..., 100. (Recall that 0 is a multiple of 3, which is why 1 goes in this group.) The remaining balls must have numbers that are 1 LESS than a multiple of 3: 2, 5, 8, 11, 14, ..., 98. We can color them red. So now, every number has a color, right? If you can figure out which color combinations will give you a multiple of 3, then your problem becomes one about selecting colored balls out of a jar... which is probably more like the problems you've dealt with in the past, right? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ Date: 11/05/2009 at 06:15:11 From: Sepha Subject: Thank you (Probabilities divisible by three) Thank you very much for your help! I was expecting that I would just get an answer, but now that you've explained it, I know how to actually do it! Thank you! |
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