Number 9 is unique. 1x9= 9\n                    2x9=18= 1+8 = 9\n                    3x9=27= 2+7 = 9\n                    4x9=36= 3+6 = 9\n                    5x9=45= 4+5 = 9\n                    6x9=54= 5+4 = 9\n                    7x9=63= 6+3 = 9\n                    8x9=72= 7+2 = 9\n                    9x9=81= 8+1 = 9\n                    9x10=90=9+0 = 9
202066
you know... I don't think that's very fair... You asked a question and then answered it. So now I must think of something else that is special about number 9...\n\nYou can multiply by 9 on your fingers and always get the right answer. For example, say you want to multiply 4 and 9. Holding all ten fingers up, put down the fourth finger from the left. Tada! 3 fingers are on the left of the finger you put down, and 6 fingers are on the right. the answer? 36! Just put down the finger corresponding to the number you are multiplying by 9.
80371
The answer to life is 42.\n\nThe question to life, however is 6x9. You can prove that 6x9 is 42 if you multiply the numbers in base 13. Also you can translate 6x9 to mean 1+5X8+1.
231359
1+1+1+1=4\n2+2=4\n4x1=4\n2x2=4\n3+1=4\n1+3=4\n1x4=4\nHope that answer your question, and u could count by your fingers or u can get objects to help u
529089
9+6         6+69          69+60\n9-6          6-9             69-6\n9+69       6-69           60-9\n9-69        6+60          60-6\n9-60        6-60           \n\n60/6\n\ncan you go so far as to do (60+6)/6\nif so... its a very long list... \nbut 9*6 is the only outright multiplication one\n\ncan you have negative #s as well thatare 2-digit?
51556
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49\n\n6 f the above numbers
124372
First fact: L times L equal to a number ending with N.\n\nL = 0 1 2 3 4 5 6 7 8 9\nN = 0 1 4 9 6 5 6 9 4 1\n\nSince L is not the same as N, it would eliminate a few from the above and leave only:\n\nL = 2 3 4 7 8 9\nN = 4 9 6 9 4 1\n\nFact 2: L times N is 9 or less; otherwise, the product would be a five-digit number.\n\nThat would narrow the above down to:\n\nL = 2 9\nN = 4 1\n\nHowever, if L = 2 and N = 4, L times N would be 8, much greater than L (in product.)  So that would eliminate that combination.\n\nL = 9\nN = 1\n\nFact 3: look at the "NO" times L part, with N = 1 and L = 9.  If O is 2 or above, it won't work because that would lead to a five-digit product.  So letter O is either zero(0) or 1.  But since N is already 1, letter O = zero(0).\n\nO = 0\n\nEnding: putting things together\n\n10E9\nx  9\n----\n9E01\n\nThe only E making sense here is 8 (9 x 8 + 8 = 80)\n\nSo the answer is\n\n1089 x 9 = 9801
646746
=> 14x -8+5-9x=1-3x\n=> 14x-9x+3x=1+8-5\n=> 8x=4\n=> x=4/8=1/2=0.5\n\nCHEERS!
439618
9 obritals:\n\n(3, 0, 0)\n(3, 1, -1)\n(3, 1, 0)\n(3, 1, +1)\n(3, 2, -2)\n(3, 2, -1)\n(3, 2, 0)\n(3, 2, +1)\n(3, 2, +2)
461519
just divide the inside by a perfect square (i.e. 4,9,16) and write it like this\n(this is the first problem you have)\nV(3X9) because 3X9=27\ntake the square root of the perfect square and move it to the outside\n3V3\nwhen you already have a number outside the radical, multiply it by the number you just simplified from the inside of the radical\n3V32\n3V(16X2)\n3X4V2\n12V2
650171
The equation would look like:\n9 * (2X - 3) = X - 4 so \n18X - 27 = X - 4\n17X - 27 = -4\n17X = 23\nX = 23/17\nX = 1.35
469623
So call the number\nB = {b_0,b_1,b_2,b_3,b_4,b_5,b_6,b_7,b_8,b_9}\n(this reverse indexing makes things easier).\n\nYou must have b_0 != 0, since you were told it was a ten-digit number. So b_0 is between 1 and 9.\nSo b_0 = 0 - (number of the b_i which are non-zero)\n\nAlso note that Sum_i=0..9 (b_i) = 10 since B has ten digits, and each b_i has to have a single value between 0..9\nHence Sum_i=1..9 (b_i) = (10 - b_0) for any particular choice of b_0\n\nTo prove the number is unique, see below:\n\nIt's easy to show 6210001000 is one possible answer.\nb_0 = 6 since there are six 0's in B\nb_1 = 2 since b_2, b_6 =1\nb_6 = 1 since b_0 = 6\nAll other b_i are 0 since there are no other non-zero b_i.\n\nTo prove the number is unique, the easiest way seems to be to show that the other b_0 between 1..5 and 7..9 are impossible by (proof by contradiction).\n\nb_0 = 9 is impossible since then b_9 must be 1, which gives a contradiction giving '9000000001', which only has 8 zeros, also b_1 must be 1, et.c\n\nb_0 = 8 gives '8000000010', but then b_1 must be 1, and it has one less zero, so that becomes '7100000100'...\nbut that has two ones, so b_1++, and b_2 must be 1, and you have again one less zero so b_0=6, giving '6210001000', which finally works.\n\nHarder to show that b_0 = 1..5 are impossible.\n\nLet's start from the opposite extreme, b_0=1.\nThen Sum_i=1..9 (b_i) = (10 - b_0) = 9\nThis constraint means each b_i can only be 0, 1 or 2.\nSince we set b_0=1, this means we have one zero and eight ones e.g. B='11111111110', doesn't matter where you put the 0. But that doesn't work since clearly b_1 must >=8, thus Sum_i=1..9 (b_i) = 8+7>9 which is a contradiction.\n\nIt should be trivial to crank out the cases b_0=2..4, maybe there is a more compact proof.
377969
The numbers for this particular answer are 9 and 10.\n\n9 + 10*3\n9 + 30\n=39
438283
x^2-6x+9 = (x-3)(x-3)\nx^2-9 = (x-3)(x+3)\n\nMultiple everything by (x-3)(x-3)(x+3)\n\n(x+3)(x+3) - (x+2)(x-3) - 5(x-3)(x+3)\n\nx^2+6x+9 - (x^2-x-6) - (5x^2-45)\n\nx^2+6x+9-x^2+x+6-5x^2+45\n\n-5x^2 + 7x + 60
615183
1, 3, 9, 27.\n\nThese are the powers of 3 if you noticed.\n\n2 = 3-1\n4 = 3+1\n5 = 9 -3-1\netc.
531850
I think it's 9*4 - (9+4) = 36 - 13 = 23.
612704
here is one\n024331\nassuming that the 2&3 numbers are the only evens\nassuming only one one\nassuming  2&3 numbers are not same numbers\nassuming only one set of repeating numbers\nalso 024551, 024771, 024991, 026331, 026551, 026771, 026991, 028331, 028551, 028771, 028991, 042331, 042551, 042771, 042991, 046331, 046551, 046771, 046991, and so on.
127108
9^8 is 9 times itself 8 times\n9^8\n= 9*9*9*9*9*9*9*9*9\n= 43046721