You don't have to see any lost ^^.. It's more like uhm.. well.. pay attention to the 4th pic and then to his reaction ^^.. Lost: The game.. :badgrin: *Yes we still do that around here xD*
For some reason it won't load here. Click and hold the 'Image' thingy and move it to your internet tabs on top on your screen, then release it ^^ After that you'll be able to see it here too. Btw, that's a brilliant idea @ the pic xD
That's a rubbish picture considering every outcome is 9. Also was that Shotgun loaded on that episode?
That's the big joke about that picture, if you can explain to how that mathematically is possible I would appreciate that a lot. I don't know what top gear episode it is from, so I can't tell.
Considering x is the number you choose : x -> 3x -> 3x+3 -> 9x+9 so you already see there's something with nine coming ^^' then what results can you get with x being from 1 to 9 : 18 27 36 45 54 63 72 81 90 every result is nine*something, which is logical because the math is 9*somethinx+9. Something else you can notice : left digit are numbers from 1 to 9, and left digits are 8 to 0. And then uh... I was going to say something smart but I forgot.
No fail, its absolutly right 3(3x+3) = 9x + 9 Pretty strange I picked my favorite number between 1-9 and that is 3,14 but I never get my favorite movie and that is 'The Joy of Anal Sex With A Goat'
i know, it's just that i didn't finish my explanation because i forgot what i wanted to explain... xD
maybe this : "9 | x" reads as "9 divides x" and is the same as saying "x is divisible by 9" or "x/9 is an integer." As for the first question, it's a well-known trick that summing a number's digits will tell if it's divisible by 3 (if they add to a mult. of 3) and by 9 (if they add to a mult. of 9). let's prove that for x a multiple of 9, it's digits sum to a multiple of 9: So we can express x as a sum of powers of 10: x = a + 10b + 100c + 1000d + ... (for a,b,c,d single digit ints) Prove: 9 | a+b+c+d+... (9 divides the sum of the digits) x = a + (9b+b) + (99c+c) + (999d+d) + ... (equivalent to above) = (9b+99c+999d+...) + (a+b+c+d+...) (rearrange) = 9*(b+11c+111d+...) + (a+b+c+d+...) Now, we remember that x | (y+z) if x|y and x|z and see the first term is divisible by 9. We must show that (a+b+c+...) is divisible by 9: (a+b+c+d+...) = x - 9*(b+11c+111d+...) (rearrange terms) Remember, x = 9*i (x a multiple of 9) ==> (a+b+...) = 9*i - 9*(blah) = 9*(i-blah) = 9*(some int) ==> (a+b+...) = sum of digits = multiple of 9 Spoiler This topic just became unfunny