riddle 5...solved by finn...riddle 6...solved by C_stripes...

[C_stripes]

You have to take the cds out of the box. 1 out of one box, 2 out of the next and so on, then you calculate the weight and you can figure how what box has the heavy ones by doing the math. I still don't understand the 55 though.

[captnsim]

SO close...can you put it together??

[C_stripes]

Got the 55, 1 out of box 1. 2 out of box 2. etc... but I don't know where to go from there because the math I am doing doesn't add up.

55 is the amount of CD's that will be on the scale. So now its all times and divide, I just don't know where to go.

[captnsim]

Even closer...You basically have it...All your trying for is the formula to find the box...The box can be any one of the ten...

[Finn]

Alright I might be off *darn thing occupied most of my time during the orientation to one of my classes. I was asking my friends if they have ever heard it before...also stumped my engineering teacher with it  * but I think Troy is on the right track but with the wrong numbers. You put the CDs on the scale but with a set number number from each box, hence the 1 through 10 for clarification right?
We can calculate the weight of all the A CDs which comes out too...

*I just went back to copy it and work out the math but c_stripes pretty much posted my idea up*

[C_stripes]

I have 715 total grams if they were all 13 grams. so now I just need to figure out how to calculate it so that I can find out where the 15 gram ones are.

Help?

[captnsim]

you all are sooo close....

[Finn]

Ah since its still open you put the cds on the scale like C_stripes said, 55 cds in total.
W/e the total weight is on that (the 1 dollar part) you can subtract the weight of the total of the A CDs and then divide that by the difference in the weight between A and B CDs which I think gives you a number that corresponds to to however many you took out of box so and so, which would be the box of the heavy CDs.


PHEW.

??

[C_stripes]

Duh!! There really is no way FOR US to know the box. But the way to find out is to take the total amount and subtract the 715 from it.

So if the scale said 719, you subtract 715 and come out with four. Which means that box two has the cd's.  

Am I right??

[Finn]

I believe so but my brain hurts a bit lol.

[captnsim]

Congrats, C_stripes...nice thinking...

[C_stripes]

Crazy,  Now I need to do my homework. your riddles keep distracting me.  lol

[resq302]

    aarrrrgggghhh.  all this thinking is making my brain hurt.   (and I still don't get it)

[captnsim]

I figured I'd post the whole answer for this one in one post.

You can determine which box has the "d" CDs in it by using the scale just one time. Here's how:

Label the boxes 1 through 10. Put one CD from box number one on the scale. Then, on top of that, put two CDs from box number two. Then put three from box number three on those, four from box four, and so on.

There are now 55 CDs on the scale. Pay the dollar and see what the total weight is. If they were all 13 grams, the total weight would be 715 grams (55 x 13). However, you know that one or more of the CDs weighs 15 grams.

Subtract 715 grams from the total weight, and this gives you the "extra weight" for the heavier CDs. Since you know that they are each two grams heavier than the others, you can divide this excess weight by two, and the result tells you how many of the heavier Cd's are on the scale, and this number tells you which box contains the "d" CDs.

For example, if the weight is 727 grams, you would subtract 715 from this, leaving you with 12 grams of "excess weight." Divide this by 2, and you know that there are 6 of the heavier CDs on the scale. Since the number of CDs from each box coincides with the number assigned to that box, you now know that box number six has the "d" CDs.