look back in my question 6=5=9 okay just a little hint I hate making people look dumb but thatsd the way it looks right now, because most of you got 11 I;m like what the  hell okay i will give you the frist answer 6+5=NINE, because you you draw six lines and it and then fill in the rest and its spells out NINE okay see what the other spells out if you don't know it is 5+4
134651
so lets see....you are suppose to draw six vertical lines....\n|  |  |  |  |  |\nthen if you want to turn those lines into the word nine, you have to draw in five more lines, thus 6 + 5 = NINE\n|&#92;| | |&#92;| |E\n N  I  N  E
326491
first you figure out the number of possible outcomes you can have. which would be 6x6 = 36. \nThen you figure out how many combinations you can make to make it greater than 7:\n5 and 3\n5 and 4\n5 and 5\n4 and 4\n4 and 5\n3 and 5\nwhich is 6.\n6 over 36 is one sixth. therefore your answer is 1/6.
674426
Okay I got the answer 3121 from Scott, but I had a mathematical strategy.\n\nI narrowed down the possibilities to all multiples of 5 times a number with numbers digit of 4 or 9 plus 1.\n\nThis is because to keep the chain of divide by 5 subtract 1 remainder going, we would start with a number x ending in either 4 or 6. We subtract x/5 and 1, so we are basically subtracting the whatever factor times 5 yields the closest muliple. The result from x-(x/5)-1 has to end in either 1 or 6, so that we have a multiple of 5 with remainder 1 and can keep the sequence going. Now if it ends in 1, the factor was even. Ends in 6, factor was odd. If factor was even, we need to subtract the factor plus one, an odd number, to get another number ending in either 1 or 6. If we subtract a number ending in 5, we get a number ending in 6. Therefore, if the factor is even it has to end in 4. And for numbers ending in 6, we have to subtract an even number (odd factor + 1)subtract a number ending in 0, so only factors ending in 9 work.\n\nI was then testing every factor ending in 4 or 9. Take the factor, multiply by 5, add one, subtract (factor + 1), see if the new number minus one divided by 5 yields a factor ending in either 4 or 9.\n\nI was up to 149 when I decided to refresh the page and saw the answer.  Jeez I would've had to go up to 624. -.-;;\n\nAs it is, I only got a chain of 3 which didn't end in a multiple of 5.\n\nSo if I had continued testing throughout the night I would've gotten it. :)
128393
well we know the 3 space must be even, in order to avoid the last space resulting in another 5\n\nhmmm... i'l work on it, provided you're working as well!\n\n..and #3 mustn't be 0, to avoid another 0\n..lol i'd have had it if i had been home...i made a program to do it, and then left before i finished/ran it lol\n\nbut that, is of course, the correct asnwer, 396*45=17820\n\nhaha thanx for the challenge!
66767
A=B\nA-B=0\nDividing by A-B, which is zero, is not allowed\nSince 5/0= 6/0 = 7/0= infinity\nthis doesn't mean that 5=6=7
210255
Can we only use the number once?\n\n9+5+26-26=14\n\n9*5-26-5=14
528970
Add them... 8+7+7+5+3+2+2= 34 and divided by 7\nYyou have 34/7=\nYou know 4x7= 28 so you need 6 more to reach 34\nso you have 4 and 6/7
53649
Yahoo's formatting messes this up, but if you cut-and-paste it into Word or Notepad, then use font Courier New it should make more sense.\n\nFirst question\n\n  1      1     11\n---- + ---- = ----\n 3x      2     12\n\n  4       6x     11x\n----- + ----- = -----\n 12x     12x     12x\n\n\n4 + 6x = 11x\n  - 6x = -6x\n-------------\n4      =  5x\n\ndivide both sides by 5\n\n 4\n--- = x = First answer\n 5\n\n\n******\nSecond question\n\n1.5x + 9 =  4.5\n     - 9 = -9.0\n----------------\n1.5x     = -4.5\n\ndivide both sides by '1.5'\n\nx        = (-4.5)/(1.5) = (-3)\n\nx = -3 = Second answer
582380
shouldn't it be 4, 8, 15, 16, 23, 42
629274
actually, i think it's 15 because however much you add or subtract in the first two, it always adds up to 6.\nAND\nyou always add once and subtract once\nTHUS\nyou have to subtract 3 from 18
79232
its impossible because there needs to be either one less sign or one more number look:\n\nin a the order:\n2 * 4 - 6 + 8 /\n              ^there is an extra sign!!
417645
If the number begins with 5, there are still three of the odd digits available to fill the last place.  If a 3 is used for the last place, the others are filled using 1, 2, 3, 4, and so there are 4! arrangements.\nIf 1 is used for the last place, that leaves 2, 3, 3, 4 for the others, and so the number of arrangements is  4!/2! because of the repeated 3.  So that gives \n4! + 4!/2! = 36\n\nIf it begins with a 3, the last place can be filled with 1, 3, or 5, and  the numbers left to fill the remaining four places are all different and so the number of possibilities is\n3*4! =72\n\nIf it begins with a 2 or 4 ( two choices), and the last with 1 or 5 (two choices), there are 4!/2! ways of filling the other places because 3 is repeated.\n2*2*4!/2! = 48\n\nIf it begins with 2 or 4 and ends with 3, then the other four digits are all different, and the number of possibilities is\n2*4! = 48\n\nAdding all these together gives 204.  Yes.
380873
This is how you would solve this.  Make all of the numbers the same size by adding zeros in front of the numbers that are smaller.  Than you add column by column starting with the last column.  \n\nSo, first we take 7 + 6 + 5 + 5 + 3.  That equals 26.  We put a 6 on the right and carry a 2 to the top of the previous column.\n\nNext we take 6 + 5 + 4 + 6 + 3 and get 24.  We have to add the 2 we carried and get 26.  We put a 6 at the bottom of the column and carry a 2 to the previous column.\n\nNext we add 0+0+0+0+3 and get 3.  We need to add 2 to that for what we carried over.  We get 5.\n\nNext we add 0+0+0+9+3 and get 12.  We didn't carry over from last time, so we get 12.  We put a 2 at the bottom of the column and carry a 1 to the previous column.\n\nContinue to do the same thing for all columns.  \n\n0000111022 carry numbers\n\n00000000067\n00000000056\n00000000045\n00000789065\n33333333333\n____________\n\n33334122566
237115
ok, let's start with the biggest number:\n\n18: 18 can only go with 7 because the nearest perfect square is 25 and other number can put it to 36.\n\nsimilarly, for 17 and 16. they can only go with 8 and 9.\n\nso, so far, we have\n\n18 + 7\n17 + 8\n16 + 9\n\nnow, 2 can only go with either 7 or 14, but since 7 is taken, 2 must go with 14\n\nnow, 11 can only go with 5 or 14, but since 14 is taken, 11 must 5.\n\nnow, 4 can only go with 5 or 12, but since 5 is taken, 4 must 12.\n\nnow, 13 can only go with 3 or 12, but since 12 is taken, 13 must 3.\n\nnow, 6 can only go with 3 or 10, but since 3 is taken, 6 must 10.\n\nnow, 15 can only go with 1 or 10, but since 10 is taken, 15 must 1.\n\nnow, 10 can only go with 6 or 15, but since 15 is taken, 10 must 6.\n\nfor 3, it can go with either 1 or 6 or 13, but 6 and 13 is taken, we only have 1 left.\n\nso....sally goes with 3\n\nin summary\n\n18+7\n17+8\n16+9\n15+1\n14+2\n13+3\n12+4\n11+5\n10+6\n\n\nso...lets summarize what we have\n\n18 + 7\n17 + 8\n16 + 9\n2 + 14\n13 + 12\n11 + 5\n14 + 7
119811
well lets take the first example- ok so the middle number is 11. and the number with no 'x' in it is 30. what makes 30 = 6 x 5 or 3 x 10. so try adding and subtracting those numbers in your mind: 6+5=11 ... 3+10=13 ...6-5=1...5-6=-1...10-3=7...3-10=-7. so which one of those give 11  ->> 6 + 5. \nso all u do is x^2+ 6x + 5x + 30 [split 11 as 6 + 5]\nur next step ; take 'x' common for the first to terms and 5 common for the 2nd two terms, u get\nx(x+6)+ 5(x+6)\ntake the bracket term multiplied by the un-bracket terms\n(x+5)(x+6)\n\nvoila...problem solved
240516
Actually, the above answers are incorrect.\n\nWe have 5 cards and 4 suits.\n\nLet's, say we draw the cards in the following order:\nAce, 2,3,4,5\n\nThere are 4 suits. So there 4 ways we could draw an Ace first; 4 ways we could draw a 2, 2nd, etc. And so there are 4x4x4x4x4 = 1024 ways of drawing the cards in the above order. However the cards can be drawn in any order, so we have to multipy 1024 by the number of ways the five cards  could be ordered when they are drawn. The number of ways the 5 cards could be ordered is 5 factorial (i.e 5!) = 5x4x3x2x1=120. So the number of ways of drawing a hand consisting of the 5 specified cards, in any suit and in any order  =  1024x120 = 122,880
225599
this really only works when its done verbally...when people are looking at the numbers, its much easier to realize that the answer is 4100 instead of 5000, pretty neat one isnt it