605
   

NeoPets Riddles (Lenny Conundrums) and Answers Here

 
 
godmother01
 
  1  
Reply Wed 28 Jan, 2009 09:35 pm
@myssinu,
I think the answer is 598925 [this is what my husband came up with]. Please let me know if I'm right.
edstock
 
  1  
Reply Wed 28 Jan, 2009 09:40 pm
@godmother01,
godmother01 wrote:
I think the answer is 598925 [this is what my husband came up with]. Please let me know if I'm right.


That is barely more than the sum of the areas of the circles, so I doubt it.
0 Replies
 
rodrunner
 
  1  
Reply Wed 28 Jan, 2009 09:43 pm
@markr,
i agree with 623671.
0 Replies
 
Nocturnaly
 
  1  
Reply Wed 28 Jan, 2009 09:45 pm
I personally got 665903 (rounded up to the nearest whole number).
I found out the circumfrence of the whole "shape". Then divided that by 4 to get 816.02869177. Times that by itsewlf to get the area of a square or equivilent to that "shape".
godmother01
 
  1  
Reply Wed 28 Jan, 2009 09:46 pm
@Conumdrumgold,
So the answer is 623671?
0 Replies
 
Cre
 
  1  
Reply Wed 28 Jan, 2009 09:47 pm
Same here (623671), but I missed adding in one of the circles the first time and sent with the wrong answer..
0 Replies
 
edstock
 
  1  
Reply Wed 28 Jan, 2009 09:48 pm
@Nocturnaly,
Nocturnaly wrote:
I personally got 665903 (rounded up to the nearest whole number).

I found out the circumfrence of the whole "shape". Then divided that by 4 to get 816.02869177. Times that by itsewlf to get the area of a square or equivilent to that "shape".


I'm afraid that isn't going to work. For a simple example, look at square with sides of length 4 and an area of 16. Then look at a 5x3 rectangle. Same circumference of 16 as the square, but only an area of 15. You might get an estimate that way, but there isn't much chance of it being right in the sixth significant digit.
0 Replies
 
polarsurf
 
  1  
Reply Wed 28 Jan, 2009 09:49 pm
well I submitted 623671. If that's not correct, then it's going beyond my skills x_x
0 Replies
 
Nocturnaly
 
  1  
Reply Wed 28 Jan, 2009 09:52 pm
Good point, Im out of this one then. I had that spark summed it out and submitted without actually thinking lol...
0 Replies
 
alyse
 
  1  
Reply Wed 28 Jan, 2009 09:55 pm
I got the same answer 623671, but I'm -no- mathematician. Hopefully since many others got the answer, it will be right.

But I know I had a VERY hard time with the area between the circles. I stumbled through the equations like a caveman. Could someone please please -please- link to a website with some background on the equations and maybe a little more information so I can learn them. Even though they will probably never come up again, I want to learn the theory behind them instead of just stumbling through numbers.
polarsurf
 
  1  
Reply Wed 28 Jan, 2009 10:01 pm
I'm no mathematician either and I'm just in 10th grade. TNT can't expect us to be some freaking Einsteins right? Haha. Anyway, if we fail, there are still many LCs to come... Does anyone know if we can change our profile picture here?
0 Replies
 
edstock
 
  1  
Reply Wed 28 Jan, 2009 10:02 pm
@alyse,
alyse wrote:
Could someone please please -please- link to a website with some background on the equations and maybe a little more information so I can learn them. Even though they will probably never come up again, I want to learn the theory behind them instead of just stumbling through numbers.


Wikipedia has what appear to be pretty good articles on both Heron's Formula and the Law of Cosines, including multiple proofs of each. Just search for either of them by name. The only other formula involved is the area of a sector of a circle, which is (angle in radians divided by 2) * (radius squared). For an entire circle, the angle is 2*pi, so it becomes the familiar pi r-squared.
brez
 
  1  
Reply Thu 29 Jan, 2009 03:17 am
@edstock,
I don't know how to explain it, so i made a picture, llike that you only have to calculate triangles and circle segments :-)

http://brezendrache.beepworld.de/photoalbum/164658/1644964_l.jpg
0 Replies
 
PeckyP
 
  1  
Reply Thu 29 Jan, 2009 05:29 am
It looks like the people who are aware of how to do it have gotten the idea right for the most part. I haven't actually done it yet. Turbine Inc, just had to distract me with their events... but I'll be doing it on my cross-province bus ride today, because the prize is so unappealing.

I should note, I will be using a different method, so it will verify other answers if one of them is right. My method will be with Calculus integrations to solve for the areas of each triangular section. I've done it before in an assignment, so shouldn't be anything new.

- PeckyP -
BrixenIvy
 
  1  
Reply Thu 29 Jan, 2009 09:30 am
it's been about 25 years since i studied trigonometry lol so i had to do a bit of memory refreshment
i came up with 623671.095 before rounding
i'm glad to see that a few other people got the same thing
0 Replies
 
markr
 
  1  
Reply Thu 29 Jan, 2009 09:01 pm
@PeckyP,
I'll be on pins an needles waiting to find out if I got it right for more than "the most part."
0 Replies
 
remedy1502
 
  1  
Reply Thu 29 Jan, 2009 10:22 pm
I got 623571... Off by 100 than the rest of you. Hmm... Gonna check my calculations tomorrow.
0 Replies
 
PeckyP
 
  1  
Reply Fri 30 Jan, 2009 02:08 pm
I got 623671.094895 as my final answer. My integrations weren't 100% accurate though. Would have a couple thousandths or tenthousandths off here and there after the definite integrals. Mind you I was doing this all in my head, so I'm just relieved to see that my answer is quite close to what was previously posted. =P

- PeckyP -
Jen Aside
 
  1  
Reply Wed 4 Feb, 2009 05:05 pm
http://images.neopets.com/games/conundrum/297_roundnround.gif

4 12 7 3 11 5 7 8 8 8 4 7 11 6 10 8 5 8 6 7 1 2 13 5 6
0 Replies
 
edstock
 
  1  
Reply Wed 4 Feb, 2009 05:09 pm
Well, the obvious thing to do is to start at some point on the circle, move the number of spaces around the circle given by each number, and find the question they want us to answer.

So, where to start, which direction to move and whether we move continuously in the same direction or change direction in some pattern are the subquestions.
 

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