Post your riddles and math/logic problems here
riddle:
Which word in the English language becomes shorter when it is lengthened?
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Post your riddles and math/logic problems here
riddle:
Which word in the English language becomes shorter when it is lengthened?
short
Yea, you got any?
another riddle:
Must Get Here Must Get Here Must Get Here
3 Mustketeers?
I dont know any riddles.
A barrrel weighs 20 pounds. A man puts something in it, and now it weighs less than 20 pounds What did he put in it.
A burning match.
That good be, but not the answer I'm looking for
Oh, I know!
An anti-gravity field modulator!
No
Hint;
Starts with an H
He but I suppose you would need alot of that for 20lbs.
a hole. duh
if a it took a man 5 hours to dig 14 holes, how long would it take him to dig half of a hole?
you can't dig half a hole.
approximatly 0.18 hours
really
Like Ken said, you can't dig half a hole
why not
As soon as you take some dirt out it's a whole hole
yup, that's the right answer ken and drive.
See, I'm not completly stupid.
Riddle: (easy one)
The Pope has it but he does not use it. Your father has it but your mother uses it. Nuns do not need it. Your lady friend's husband has it and she uses it. What is it?
a penis.Quote:
Originally posted by Driveway
See, I'm not completly stupid.
Riddle: (easy one)
The Pope has it but he does not use it. Your father has it but your mother uses it. Nuns do not need it. Your lady friend's husband has it and she uses it. What is it?
Better than the real answer, but no
My riddle: (I gave up on the other one)
A hunter sets out from camp at 4 o'clock in the morning with a compass. According to the compass, he travels due south for 4 hours. At that time he sees a bear and kills it (he is a hunter after all). He drags the bear with him for 4 hours due east, and he takes a quick break. Then he drags the bear due north for 4 hours at which time he arrives back at his camp where he decides he doesn't want the bear after all, packs his SUV and leaves.
So what color was the bear that he killed?
white
What has been done by every male, except for one? Normally does it with many, but can do it by herself. Is world-renowned for her infinite skill and great noise. And gives many great pleasure.
P.S. You'll probably never get this.
Well here's a maths one (quite easy):
By considering the series,
1 + t + t^2 + t^3 + ... + t^n
sum the series,
1 + 2t + 3t^2 + 4t^3 + ... + nt^(n-1)
for t not = 1, where a^b denotes a raised to the bth power.
Have fun!
Good problem!
I think it's
(n * t^(n+1) - (n + 1) * t^n + 1) / (t-1)^2
I even sort of proved it.
Yeah that's correct, but where's the proof??Quote:
Originally posted by Nick
Good problem!
I think it's
(n * t^(n+1) - (n + 1) * t^n + 1) / (t-1)^2
I even sort of proved it.
This is how I found the answer
This I proved this was equal toCode:1 + t + t^2 + t^3 + ... + t^n
+ t + t^2 + t^3 + .... + t^n
+ t^2 + t^3 + ... + t^n
+ t^3 + ... + t^n
....
+ t^n
1 + 2t + 3t^2 + 4t^3 + ... + nt^(n-1)
After that I solved the sums and then used induction
to prove/check the final answer.
I see, well done, but I have a much shorter proof, using 1 + t + t^2 + t^3 + ... + t^n. Can you/anyone else find it?Quote:
Originally posted by Nick
This is how I found the answer
This I proved this was equal toCode:1 + t + t^2 + t^3 + ... + t^n
+ t + t^2 + t^3 + .... + t^n
+ t^2 + t^3 + ... + t^n
+ t^3 + ... + t^n
....
+ t^n
1 + 2t + 3t^2 + 4t^3 + ... + nt^(n-1)
After that I solved the sums and then used induction
to prove/check the final answer.
You'll kick yourselves...
Good Luck!
I noticed that derivative
of 1 + t + t^2 + t^3 + ... + t^n
is the 0 + 1 + 2t + 3t^2 +... + nt^(n-1)
I'm sure you could give a prove or solve it some how
using this?
Exactly! Well done.Quote:
Originally posted by Nick
I noticed that derivative
of 1 + t + t^2 + t^3 + ... + t^n
is the 0 + 1 + 2t + 3t^2 +... + nt^(n-1)
I'm sure you could give a prove or solve it some how
using this?
If you sum the first series, using the standard geometric series result, you can then differentiate that result to get the sum of the second series, which is the same answer as you got.
Like this:
Summing first series (using standard geometric series result):
1 + t + t^2 + ... + t^n = (1 - t^(n+1))/(1 - t)
d(1 + t + t^2 + ... + t^n)/dt = 1 + 2t + 3t^2 + ... + nt^(n - 1)
Which implies that:
1 + 2t + 3t^2 + ... + nt^(n - 1) = d((1 - t^(n+1))/(1 - t))/dt
= (1 - t^n + nt^(n + 1) - nt^n)/((1 - t)^2)
I think that's quite a nice little solution...
If you havn't got it already check out
concrete mathematic --- A foundation for computer science.
Lots of neat problems, most of them are solved in the back
of book.
I'm trying to read this book in my spare time, but I have
lots of course work so I havn't gotten too far. It's tough
to read.