The worlds first riddle!

Yeah, I too did not like that last puzzle, so how about:


My first three and last three are the same, and many people stay with me when they die. Who am I

(You're looking for an 11-letter word.)



Annabella is playing the videogame Space Critters, and she finally loses her last turn after several hours (and after killing numerous invading space critters).

She was rather surprised to discover that she had scored the maximum number of points she could have while averaging exactly 9975 points per turn.

If in Space Critters you start with 3 turns and you earn an extra turn with every 10000 points you score (e.g. you earn an extra turn at 10000, another at 20000, another at 30000 etc.), what was Annabella's final score

[size=8]SPACE CRITTERS
N = number of 10000 point games she played (she scored 0 on the last game)

10000 * N / (N+1) = 9975
N = 9975/25
N = 399

She scored 3,990,000 points.[/size]

You've done the candy puzzle at least once before. I didn't get it then.

Candy puzzle

I seem to remember something along those lines, but I don't fully understand this one even with the answer. What do you make of it?

Let D(n) be the number of different flavors that can be made with n candies, and let E(n) be the number of different flavors that can be made with n candies, but not with any smaller number.

The number we are seeking is F = E(1) + E(2) + ... + E(10). Note that D(n) = Sum E(d), where the sum is over all d such that d divides n. This gives us that D(6) = E(1) + E(2) + E(3) + E(6), D(7) = E(1) + E(7), D(8) = E(1) + E(2) + E(4) + E(8), D(9) = E(1) + E(3) + E(9), D(10) = E(1) + E(2) + E(5) + E(10). Thus, we can express F as D(6) + D(7) + D(8) + D(9) + D(10) - 4*E(1) - 2*E(2) - E(3) = D(6) + D(7) + D(8) + D(9) + D(10) - D(3) - 2*D(2) - D(1).

To find what D(n) is, imagine we have 4 sticks and n flavorless candies. Arrange the sticks and candies in a straight line; this can be done in (n+4) choose 4 = (n+4)!/(n!4!) ways. Note that each such arrangement corresponds to a different flavor by the simple mapping of letting all the candies to the left of the first stick be apple, the candies between the first and second sticks be banana, etc. This is a standard discrete math technique.

It follows that F = 14C4 + 13C4 + 12C4 + 11C4 + 10C4 - 7C4 - 2*6C4 - 5C4 = 2681.



BTW: SPACE CRITTERS

May I respectfully suggest you re-read the question, I think you will find that total is far to low.

Hmmm. I guess I don't fully understand the problem.

Well, here goes - Space Critters:

Let T be the total number of turns Annabella had (including the initial 3), and let S be her final score. Then by the rules of the game we have that S = 9975*T and T = 3 + floor[S/10000], where floor[x] is the greatest integer less than or equal to x. This gives that T = 3 + floor[9975*T/10000] = 3 + floor[(10000*T-25*T)/10000] = 3 + T + floor[-25*T/10000] = 3 + T - ceil[T/400], where ceil[x] is the smallest integer greater than or equal to x. We therefore get that ceil[T/400] = 3, and it follows that we must have 801 ... is less than or equal to ... T ... is less than or equal to ... 1200.

Since Annabella scored the maximum number of points, T = 1200, and so her final score was 9975*1200 = 11,970,000.




Thoh is taking a series of exams, and it turns out that he'll have to score a 97 on the last one in order to average 90 for the entire series. But even if he scores as low as a 73, he'll still average an 87.

How many exams were in the series

[size=8]EXAMS
Counting this one, eight.[/size]

Mark:

EXAMS
Counting this one, eight

Does anyone believe that? Well, let's see:
Let S be the sum of all of his exam scores before the final, and let there be a total of n exams. Then we have (S+97)/n = (S+73)/n + 3, or 3n = 24, and so n = 8.

Yup! Mark is indeed right again

In general, if a difference of x points on the final exam corresponds to a difference of y points in the overall average, there are x/y exams in total.



Mark wants to draw an equilateral triangle (all sides the same length), and wants to do it as neatly as possible. To that end, he's using a large piece of graph paper made up of squares, and will have each edge go between two points on the graph paper.

Can such a neat triangle be drawn

i got a weird contradicton on my paper so im gonna say no to the triangle problem

EQUILATERAL TRIANGLE
Some time back, some guy called markr offered this solution:

http://www.able2know.com/forums/a2k-post1052977.html#1052977

sugar in my coffee," Mark said sweetly.
"I'm feeling ill," Mark croaked.


Your time starts now

Thoh:

I'm gonna say no to the triangle problem

(Good call)

"Californian wood is the best," Said Mark excitedly.

[size=8]WOOD
opined?[/size]

i missed the 6 letter part haha

3 minutes
two wise guys
who's in charge
the aftermath
150 pounds

Thoh:

3 minutes
two wise guys
who's in charge
the aftermath
150 pounds


"Bob weighs 150 pounds" is the correct answer.
His sister weighs 10 pounds. Thus, Bob weighs 140 pounds more than his sister and their combined weight is 160 as per the problem requirements.
The common mistake is to quickly jump to the conclusion that "Bob weighs 140 pounds and his sister weighs 20 pounds." But, Bob would then erroneously weigh 120 pounds more than his sister. Unfortunately no one made that mistake.



Does size matter?

I have a creeper plant that is climbing up and around a cylindrical tree trunk in a helical manner. The tree trunk has a height of 480 inches and a circumference of 32 inches. If the creeper covers a vertical distance of 60 inches in one complete twist around the tree trunk, what is the total length of the creeper

[size=8]CREEPER
544 inches

sqrt(32^2 + 60^2) * 480 / 60[/size]

Mark:

CREEPER
544 inches

sqrt(32^2 + 60^2) * 480 / 60


I know a lot of you thought the question unfair because Mark is a horticulturist and knows all about square roots. Therefore, to make visualization easy, it is convenient to conceptually open out the bark of the tree trunk and flatten it. The cylindrical surface will then be a rectangle. It may be noted that:

Width of the rectangle = Circumference of the cylinder = 32 inches.

Height of the rectangle = Vertical distance on the cylinder = 60 inches (in one twist).

Using Pythagoras' Theorem for a right-angled triangle, Length of the hypotenuse = (322 + 602) 1/2 = 68 inches.

Now, the number of twists the creeper makes around the tree trunk is 8 (= 480 / 60). If the length of the creeper (as given by the hypotenuse) is 68 inches in one twist, then the total length of the creeper in 8 twists is 544 inches. (If you came up with any other answer , you were barking up the wrong tree)



Now before I go to bed, this is a real tester:

Donald Knuth, one of the most famous computer scientists in the world (and who was first published as a kid in Mad Magazine) believes that it's possible to make any positive integer by starting with a single 3 and then using some combination of the operations of factorial !, square-root sqrt(), and greatest integer [].

Note that n! = 1*2*...*n (e.g. 6!=720), and that [x] is the greatest integer less than or equal to x (e.g. [3.14]=3).

As an example, we can make 26 by [sqrt((3!)!)], since 3!=6, 6!=720, sqrt(720)=26.8, and [26.8]=26.


Now show that it's possible to make 10

[size=8]KNUTH
[sqrt([sqrt(sqrt((3!)!))]!)]

3! = 6
6! = 720
sqrt(720) = 26.83
sqrt(26.83) = 5.18
[5.18] = 5
5! = 120
sqrt(120) = 10.95
[10.95] = 10[/size]