NeoPets Riddles (Lenny Conundrums) and Answers Here

I think you guys have already moved past my basic math. . .I understand how you figured out the radius of the larger circle, but where did you get the smaller radius?

It's been ages since I did this kind of math!! I'm so lost!

We do not need any of them.

Read my first explanation again... or go to Anneska's page, whichever seems easier for you.

It's not complex maths, but just thinking.

tks and GL all

http://www.vias.org/comp_geometry/geom_circle_ring.html

Gives a really easy formula

So I'm taking a chance with
831

Hope I am right this time

The problem with that page is that it does nor explain why it is that way, but can be useful when answering the conundrum.

Good luck, everyone.

what i got:

diameter is 23.

the square circumscribed about the circle is 23x23
the diag of the square = diameter of the big circle = sqrt(2*23^2)

annullus = pi(R^2 - r^2) = 415.475628

tell me if i'm wrong...

Thank you both Kyorai and anneska for clarifying. It's early hours of the morning in my time zone, and my brain isn't quite moving fast enough yet.

The link helps because it has diagrams. yay for pictures!


I'm going to look it over a bit before I submit. Good luck to everyone!

Correct, you just have to multiply it by 2, for the coin has two sides.

LENNY
SO WHAT IS THE REAL ANSWER?

If you are not confident of the posted values, that's okay: you don't have to depend on somebody else to do all the work and give you the answer. Instead, take the information provided in the puzzler, use the formulas and reasoning explained in the previous posts, follow the instructions (or make up your own), and see what value you come up with. :wink:

Then reply, showing your work and reasoning, and give us your verdict!

Eliz.

LENNY
I'm not trying to copy someone else answer because i found a way to do it and i got 831 but im not sure i did a lot of work today to find that answer so dont think that im copying

I rounded it up to 831 as well

My side account [before they started drilling it into everyone's heads not to do contests on sides] got Gold on the first try, because I was lucky enough to see the problem very soon after it was posted and get the answer quickly. My main has only a silver.

There's no way to determine how fast you need to get the correct answer, because having the correct answer is the key. A million people could answer, but if all of them got the wrong answer, you could theoretically submit the correct answer days later and still get a gold trophy. In fact, one of the rounds [can't remember which] even said that the answer was based on how many chocolate items there were at the time of judging, NOT the time the conundrum was posted, so it was beneficial to wait a few days!

Also, there aren't always 250 winners. As stated, having the correct answer is the key, not simply answering first, so if only 27 people answer correctly [as has happened before], then only 27 prizes will be awarded. That makes the prize even more valuable when it's a prize exclusive to that contest. I remember one round was even specifically for the first TEN correct responses!

I rounded to 831 too. :/

This is how I got 831; notice the right-angle triangle on the picture. From that you get the formula:
R^2 = (23/2)^2 + r^2

Then you just insert this into the annulus formula (area of bigger circle - are of smaller circle):
pi*(11.5^2 + r^2) - pi*r^2 = pi*11.5^2 = 132,25*pi ... the multiply this by 2 for both sides

Using the provided graphic (thank you!), and letting the coin radius be "R" and the silver inset radius be "r", the Pythagorean Theorem gives us:

. . . . .r^2 + (11.5)^2 = R^2. . .*

The area of the gold-only portion of the coin is found by taking the area of the whole coin, and then subtracting the area of the silver-inset portion of the coin.

From the formula for the area A of a circle with radius r, A = (pi)(r^2), we then get:

. . . . .A_{annulus} = (pi)(R^2) - (pi)(r^2). . .**

Multiplying equation (*) through by (pi) and then solving, we get:

. . . . .(pi)(r^2) + (pi)(11.5)^2 = (pi)(R^2)

. . . . .(pi)(11.5)^2 = (pi)(R^2) - (pi)(r^2)

But equation (**) says that the right-hand side of the last equation above gives us the area of the gold portion of one side of the coin. In other words:

. . . . .A_{annulus} = (pi)(11.5)^2

Since there are two sides to the coin, the gold surface area (exclusive of the edge) is then given by:

. . . . .2(pi)(11.5)^2 = 830.95125687...

Rounding up to the nearest whole number gives an answer of 831 square millimeters.

Note: I do not know, nor do I make any particular claim regarding, whether "mm^2" or "sq mm" or some other unit should be included with the numerical value, or if the prize-winning answer is supposed to be just plain "831".

Eliz.

LENNY
i wrote 831 square millimeters

So it seems alot of people got it? Would you have said it was a easy one? I really hope I got into the top 250. I think I might have unless I made a tipo and missed it or something

Suppose you have five six-sided dice. Each die has the following letters on the faces: E, O, I, Y, K, and R.

What are the odds that, upon rolling the five dice, you would be able to spell the name of a Neopet species with some or all of the letters displayed on the dice? Assume each die can only be used once in the word; for example, if you rolled "A,A,B,C,D" you could use A twice, but B, C, and D only once.

Please submit your answer as a whole number, rounded to the nearest whole number. (i.e. if the odds are 1 in 100, submit the answer "100".)