NeoPets Riddles (Lenny Conundrums) and Answers Here

Thanks its good to know atleast someone got the same answer. But I think I submitted it too late.

quoting kelly's post:
Ok, so maybe this is the formula...

143.75-40.82=102.93
102.93 x 15 = 1543.95

143.75 is 2300/16 and is the amount of string used per revolution
40.82 is circumference

So Im thinking we should deduct the 40.82 cm per revolution since that is a given. 102.93 is (to my brain anyway) the extra string used per revolution as it is stretched down the pole.

then multiply by number of segments (twists)

I dont have a solution, I have a guess. A somewhat educated guess, but a guess nonetheless. (Sorry JJ)

Are you using the formulat: a^2 + b^2 = c^2 , Ed?

I don't rly see why you deducted 40.82....

I've got 1**7 too

Ed, i tend not to believe your answer for the following reason.
the total legnth of rope is 2300 .... and just accounting for the length that has to be used in the wrapping (~650) means the pole has to be less than 1700.
just throwing that out there.

Unity, what formula did you use?

Okay, you see the Math Forum page with the triangle, right?

This is what you get when you unroll the cylinder, like a toilet paper tube. The diagonal line, called the hypotenuse, is the string that wrapped around the cylinder, just like the indentation you see on the toilet paper tube before you unroll it.

The bottom of the triangle is 16 times around the cylinder, since we are told that the string wraps around the cylinder 16 times. The distance around the cylinder is the circumference, or PI times the diameter.

Now, the Pythagorean Theorem is a rule in math that says if we have a right triangle, we can take the square of each side and the sum of the squares of the two shorter sides (the right side and the bottom side) will equal the square of the long side (the diagonal side). This is often written as A^2 + B^2 = C^2, where ^2 means squaring (since the BBCode doesn't support superscripts).

The total height of the cylinder is just the right side of the triangle, so if I call the bottom A, the right side B and the diagonal C, then I want to calculate the square root of C^2 - A^2 to get B, the height of the cylinder.

I hope that helps.

Too late for what? I think there is probably a good chance of getting the prize this week.

Actually another place got the answer 15 minutes before us ...... and there were 49 lurkers. And I'm trying for gold since I've been stuck with silver for a while now. Definitely no gold But we'll get the prize.

1543.95 is what you would get for the height if the string wound around the pole once, then went straight down for a ways, then wound around the pole again, went straight down again, etc.

The problem with this calculation is that I'm pretty sure the string is wound diagonally down the pole. The difference is the same as walking from home plate to second base on a baseball field via first base, and walking straight across the pitcher's mound. The distance diagonally is a lot shorter and uses a lot less string, or, in comparison to this case, the distance from 1st to 2nd can be a lot longer for the same length of string.

So then did you get an answer of 5**6, Ed?

As I mentioned in my other post, when you go diagonally, you can cut off some vertical distance while you are going around.

No, the maximum possible height couldn't be more than 2300, because the string has to reach from top to bottom.

Right....**headdesk**

shouldn't we just give out the answer since it has been leached out by at least 3 forums already

warhammer -- nah. i like that (most times) this forum doesn't give out the exact submittable answer. makes people explain their logic, and do some work.

Ed -- a'ight. you've finally convinced me.

I just found a picture. It had the complete instructions and the answer. It showed exactly what ed figured out so a lot of ppl probably found that.

I agree. We've explained the reasoning in detail for anyone that wants to solve the problem, and given a masked answer to let people check their work for calculation errors.

Was it this one, as linked from Gaming Gutter's Lenny topic?

http://i15.photobucket.com/albums/a392/timsty/lennyconundrumworking.png