strange sum  riddle answers



13092011, 09:03 PM
(This post was last modified: 13092011 09:08 PM by Peterkin.)




strange sum  riddle answers
So as not to spoil riddles for those who come later or ponder longer.
I'll start off in the next available box. Anyone else who wants to give the answer to a riddle on the other thread, start a new post here with the page and title  kay? It's not the mean god I have trouble with  it's the people who worship a mean god. 

13092011, 09:49 PM




RE: strange sum  riddle answers
Shit, I screwed up and posted my answer in the first thread.......
You have just begun reading the sentence you have just finished reading. 

13092011, 09:50 PM




RE: strange sum  riddle answers
There, deleted.
My answer to the prince and princess riddle: I know I know! He says I choose this envelope, then requests that the king open the unchosen envelope to show that it says "banishment" to prove that his must say "marriage". The king would have to grant the marriage, lest he be proven a liar. You have just begun reading the sentence you have just finished reading. 

14092011, 10:31 AM




RE: strange sum  riddle answers
(13092011 09:50 PM)Stark Raving Wrote: My answer to the prince and princess riddle: Yass! In the version i read, he takes an envelope and throws it on the brazier, saying "To the Fates!" It's not the mean god I have trouble with  it's the people who worship a mean god. 

22092011, 08:07 PM




RE: strange sum  riddle answers
Solution re. irrational numbers.
Assumption: sqrt(2) = x/y where x and y are simplifed to contain no common factors, therefore x and y can't be both even. Indirect proof, leading to contradiction: sqrt(2) = x/y => 2 = (x^2) / (y^2) => x^2 = 2y^2 => x^2 is even => x is even => x = 2p from above: x^2 = 2y^2 => 4p^2 = 2y^2 => y^2 = 2p^2 => y^2 is even => y is even CONTRADICTION! with our premise. Imagine, this simple logic lead to the death of a young mathematicians  how horribly twisted a human mind can be? 

23092011, 02:45 AM




RE: strange sum  riddle answers
I heard they threw him overboard
Welcome to science. You're gonna like it here  Phil Plait Have you ever tried taking a comfort blanket away from a small child?  DLJ 

26012012, 12:45 PM




RE: strange sum  riddle answers
Solution to the x+2 +x3 = 2 equation:
The equation has no solution. Check it out in 3 regions: A. If x < 2 then x+2 < 0 therefore x+2 = (x+2) = x2 similarly x3 < 0 therefore x3 =  (x3) = x+3 so our equation in region A. becomes x2 x+3 = 2 The solution of A. is x= 1/2 which is not in the region of A. so it is not a solution. You can use the same logic for the other 2 regions: B. 2 <= x < +3 C. x >= +3 In each of these two regions you have a corresponding equation, without the absolute value: B. x+2 x+3 = 2 C. x+2+x3 =2 The solution for B is 5=2 The solution of C. is x=+3/2 (not in the region of C.) 

27012012, 02:43 AM




RE: strange sum  riddle answers
(26012012 12:45 PM)Zat Wrote: Solution to the x+2 +x3 = 2 equation: Ah, it has really no solution. The minimum answer I've found is 5, as I've mentioned, so I though it was a trick question Here's a graph to illustrate the equation  x+2 +x3 The minimum yvalue is 5 as seen from the graph, and thus there are no points in which the yvalue is 2. Great math workout after the exams Welcome to science. You're gonna like it here  Phil Plait Have you ever tried taking a comfort blanket away from a small child?  DLJ 

27012012, 05:56 AM




RE: strange sum  riddle answers
251209 = 42


27012012, 08:43 AM




RE: strange sum  riddle answers
Quote:New puzzle: Let our number be a, in which a = (x0) + (10)(x1) + (10^2)(x2) + (10^3)(x3) + .... + (10^z)(xz) Let the sum of our digits of a be b, in which b = (x0) + (x1) + (x2) + (x3) + .... + (xz) The quickest method to generate a number divisible by 9 and thus 3 from 10^k, where k is any positive integer, is by subtracting one from it. And thus, ab = 9(x1) + 99(x2) + 999(x3) + .... + (10^z  1)(xz) By factorisation, we are able to extract 9 out from the whole chain of numbers. This will show that ab is divisible by 9 and thus 3. Since we have proven that ab is divisible by 9 and thus 3, if a is divisible by 9, b has to be divisible by 9 in order for ab to be divisible by 9. The same goes for 3, if a is divisible by 3, b has to be divisible by 3 in order for ab to be divisible by 3. Tedious but rewarding Welcome to science. You're gonna like it here  Phil Plait Have you ever tried taking a comfort blanket away from a small child?  DLJ 

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