Friday riddle...
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Since you started with a maths puzzle here is another. How can this be true? let a = x a+a = a+x [add a to both sides] 2a = a+x [a+a = 2a] 2a-2x = a+x-2x [subtract 2x from both sides] 2(a-x) = a+x-2x [2a-2x = 2(a-x)] 2(a-x) = a-x [x-2x = -x] 2 = 1 [divide both sides by a-x] There is a catch!! See if you can work it out. Ant. I'm hard, yet soft.
I'm coloured, yet clear.
I'm fruity and sweet.
I'm jelly, what am I? Muse on it further, I shall return! - David Williams (Little Britain)Antony M Kancidrowski wrote: 2 = 1 [divide both sides by a-x] a = x a - x = 0 division by zero...
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Antony M Kancidrowski wrote: 2 = 1 [divide both sides by a-x] a = x a - x = 0 division by zero...
Quite. It is an oldie but I like it :-D Ant. I'm hard, yet soft.
I'm coloured, yet clear.
I'm fruity and sweet.
I'm jelly, what am I? Muse on it further, I shall return! - David Williams (Little Britain) -
Quite. It is an oldie but I like it :-D Ant. I'm hard, yet soft.
I'm coloured, yet clear.
I'm fruity and sweet.
I'm jelly, what am I? Muse on it further, I shall return! - David Williams (Little Britain)The universe would be a much more interesting place if division by zero was allowed :-)
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Since you started with a maths puzzle here is another. How can this be true? let a = x a+a = a+x [add a to both sides] 2a = a+x [a+a = 2a] 2a-2x = a+x-2x [subtract 2x from both sides] 2(a-x) = a+x-2x [2a-2x = 2(a-x)] 2(a-x) = a-x [x-2x = -x] 2 = 1 [divide both sides by a-x] There is a catch!! See if you can work it out. Ant. I'm hard, yet soft.
I'm coloured, yet clear.
I'm fruity and sweet.
I'm jelly, what am I? Muse on it further, I shall return! - David Williams (Little Britain) -
OK, i have another stupid riddle for friday.... Three guys are in a restaurant, and they think they should order something for three. Everybody gives 10$, and they call the waitress. The waitress takes the 30$ and goes to the kitchen to order a meal for 30$. Because the cook knows the three guys, he tells the waitress to give them 5$ back. She walks back to the table with the 5$, and because 5$ is not really good to divide by three, the guys tell her only to give 1$ back to each and keep 2$ as tip. And here's the riddle: Everybody has paid 10$ and got 1$ back, that makes 9$, right? 3x9$ is 27$, right? The waitress has 2$, right? 27$ + 2$ is 29$, right? :doh: ?????? Where is the missing DOLLAR ??????? :doh: :confused: Nice Weekend!!!
Olli "Ooooooh, they have the internet on computers now!"
Homer Simpson
:beer: + :java: = NULL :=> X| -
OK, i have another stupid riddle for friday.... Three guys are in a restaurant, and they think they should order something for three. Everybody gives 10$, and they call the waitress. The waitress takes the 30$ and goes to the kitchen to order a meal for 30$. Because the cook knows the three guys, he tells the waitress to give them 5$ back. She walks back to the table with the 5$, and because 5$ is not really good to divide by three, the guys tell her only to give 1$ back to each and keep 2$ as tip. And here's the riddle: Everybody has paid 10$ and got 1$ back, that makes 9$, right? 3x9$ is 27$, right? The waitress has 2$, right? 27$ + 2$ is 29$, right? :doh: ?????? Where is the missing DOLLAR ??????? :doh: :confused: Nice Weekend!!!
Olli "Ooooooh, they have the internet on computers now!"
Homer Simpson
:beer: + :java: = NULL :=> X| -
Olli wrote: ?????? Where is the missing DOLLAR ??????? I'm surprized nobody blamed taxes :rolleyes:
Собой остаться дольше...
KaЯl wrote: I'm surprized nobody blamed taxes With only 3% missing? Not here...
Have a look at my latest article about Object Prevalence with Bamboo Prevalence.
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OK, i have another stupid riddle for friday.... Three guys are in a restaurant, and they think they should order something for three. Everybody gives 10$, and they call the waitress. The waitress takes the 30$ and goes to the kitchen to order a meal for 30$. Because the cook knows the three guys, he tells the waitress to give them 5$ back. She walks back to the table with the 5$, and because 5$ is not really good to divide by three, the guys tell her only to give 1$ back to each and keep 2$ as tip. And here's the riddle: Everybody has paid 10$ and got 1$ back, that makes 9$, right? 3x9$ is 27$, right? The waitress has 2$, right? 27$ + 2$ is 29$, right? :doh: ?????? Where is the missing DOLLAR ??????? :doh: :confused: Nice Weekend!!!
Olli "Ooooooh, they have the internet on computers now!"
Homer Simpson
:beer: + :java: = NULL :=> X|The logic flaw is in the very first statement. Olli wrote: Everybody has paid 10$ and got 1$ back, that makes 9$, right? Everyone paid $10 and got back 5/3 of a dollar (5 returned by the cook divided by the 3 people). They then each contributed 2/3 of a dollar to the tip (2/3 * 3 = $2) 5/3 - 2/3 = 3/3, leaving them with with a dollar each. Marc Microsoft MVP, Visual C# MyXaml MyXaml Blog Hunt The Wumpus RealDevs.Net
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Since you started with a maths puzzle here is another. How can this be true? let a = x a+a = a+x [add a to both sides] 2a = a+x [a+a = 2a] 2a-2x = a+x-2x [subtract 2x from both sides] 2(a-x) = a+x-2x [2a-2x = 2(a-x)] 2(a-x) = a-x [x-2x = -x] 2 = 1 [divide both sides by a-x] There is a catch!! See if you can work it out. Ant. I'm hard, yet soft.
I'm coloured, yet clear.
I'm fruity and sweet.
I'm jelly, what am I? Muse on it further, I shall return! - David Williams (Little Britain)Dividing both sides by zero. I can use the same thing to prove that 1 hour is equal to one minute. See if you can do that ? :) "One of the Georges," said Psmith, "I forget which, once said that a certain number of hours' sleep a day--I cannot recall for the moment how many--made a man something, which for the time being has slipped my memory."
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You cannot divide both sides by a-x because a-x = a-a = 0. Can't divide by zero. Every programmer knows that. ;P I have a symbiotic relationship with my computer.
dacris wrote: You cannot divide both sides by a-x because a-x = a-a = 0. Can't divide by zero. Every programmer knows that. Duh! Wanna know why I'm still a hobbiest?
The only way of discovering the limits of the possible is to venture a little past them into the impossible.--Arthur C. Clark