How get 666 from nines digits 1 - 9 without repeating?
121048
I got 4 answers but there is more\n\n1 + 2 + 3 + 4 + 567 +89 = 666\n123 + 456 + 78 + 9 = 666\n9 + 87 + 6 + 543 + 21 = 666\n(975 - 864) x 1 x 2 x 3 = 666
508816
Start with the fixed part... 2.1 is 21/10.\n\nThe rest of the number is 0.036363636...\nLet's call that x.\n\nx = 0.036363636...\nNotice what happens if you multiply by 100.  I picked 100 because there are two repeating decimals.  If there were 1 you would pick 10, if 3, you would pick 1000, if 4 pick 10000, etc.\n100x = 3.6363636...\n.......x = 0.0363636...\n\nSubtract and the repeating decimals are cancelled:\n99x = 3.6\n\nMultiply by 10 to get rid of the last remaining decimal:\n990x = 36\n\nNow solve for x:\nx = 36/990\n\nNow you just need to add your two fractions:\n21/10 + 36/990\n\nGet a common denominator:\n21*99/990 + 36/990\n\n= (2079 + 36)/990\n= 2115/990\n\nNow reduce:\n= 235 / 110\n= 47 / 22\n= 2 3/22\n\nSo your answer is 2 and 3/22.\n\nYou can also do this in one step, without having to do the two parts:\n100x = 213.63636...\n.......x = .....2.13636...\n\nSubtracting:\n99x = 211.5\n\nMultiply by 10 to remove the decimal:\n990x = 2115\n\nSolve for x:\nx = 2115/990\nx = 47/22\nx = 2 3/22\n\nA final way, if you now understand the underlying math is to just do the following:\n\nPull out the part in front of the decimal (2)... just keep that for later.\n\nThe remaining part is 0.136363636... \nSince we have two repeating decimals, use two nines (99).  However, since it is offset by one fixed digit, add a trailing zero (990).\nTake the fixed digit (1) and subtract it from your number (136, ignoring anything but the first repeat).  This leaves 135.\n\nSo the fraction should be 135/990.  This reduces to 3/22.\n\nSo it becomes 2 and 3/22.\n\nAlways check that the final result makes sense.  Obviously the first part of the mixed fraction should match the decimal (2) and it does.  If you put 3/22 in your calculator, you get 0.1363636...\n\nThere's 3 different methods and they all give you 2 and 3/22.
374969
Let the number be X\nThen X = 1000a + 100b + 10c + d, where a,b,c,d are the digits of X.\nLet the 'flipped' number be Y.\nThen Y = 1000d + 100c + 10b + a\nSo, X - Y = 1000a + 100b + 10c + d - (1000d + 100c + 10b + a)\nor\nX - Y = 999a + 90b - 90c - 999d\n= 9 * (111a + 10b - 10c - 111d) .\nwhich is divisible by 9.\nThe sum of the digits of any number divisible by 9 is itself divisible by 9.*  So if you sum the digits of X-Y, you will get a number divisible by 9.  Then if you sum the digits of that number, you will get another (smaller) number divisible by 9.  If you keep doing that, you will eventually get to 9. [I assume that's what you meant.  If you sum the digits just once you wont necessarily get 9, as in your example. 6354-4536 = 1818 which sums to 18.]\n\n* You can prove that the sum of the digits of any number divisible by 9 is itself divisible by 9 by induction.\nThe first 2-digit multiple of 9 is 18, and 1+8 = 9.\nAssume it is true for k*9 where k is a natural number (1,2,3,...)\nProof for (k+1)*9:\n(k+1)*9 = k*9 + 9\nIf the last digit of k*9 is 0, then adding 9 just adds 9 to the sum of the digits of k*9.  Since the sum of the digits of k*9 is divisible by 9, so is this sum + 9.  We are done.\nIf the last digit is anything else and the tens digit is <9, adding 9 is the same as adding 1 to the tens digit and subtracting one from the one's digit.  So the sum is the same as the sum of k*9 + 1 - 1 = k*9 and we are done\nIf the last digit is anything else and the tens digit=9 then adding 1 to the tens digit means subtracting 9 from the tens digit and adding 1 to the hundred digit, which means the sum is the same as the sum of k*9 + 1 - 9 + 1 = the sum of k*9 - 9 which is divisible by 9.  (This can be extended to carrying a 1 to the other places as well.)
281312
A 9 digit number has exactly 1,000,000,000 combinations if you include the code 000-00-0000 through 999-99-9999.  So we are about 1/3 of the way to running out of numbers if we are recycling.
512123
Hmm... 370 and 371 both seem to work.\n\n* * * * * * \n\nI wrote a little program to test all the numbers from 100 to 999.  There are actually four answers:\n\n153\n370\n371\n407\n\nAlso, by "the cube of the sum of the digits", you mean each individual digit is cubed, then you add the sum of the cubes, correct?  For example, 153 is one of the answers because 1^3 + 5^3 + 3^3 = 153.  (It seems like a trivial question if you interpret it the way Dan did, since there are only five numbers to test.)
419835
42*10=420   42*100=4200\nnow u do it
397390
Write it as 999.9, and stick a line over the 9 after the decimal point.
20804
You need to round off the number to 4 digits and use the proper exponent to reflect the entire number:\n\n1.496 x 10^11  Your first answer is correct.
85006
the only way to make 1991_1991 divisible by nine is to choose _ the missing number by 5\n\n199151991 is divisible by nine\n\nso 199151991 / nine =221 2 7999 \n\nSo there is 3 numbers not 4 in the first so i say\n0221 2 7999 x nine = 1991 5 1991
477152
... 27, 26, 676, 678, 677, 458329, 458331, 458330, ...\n\nThis is sequence A059417 in the Online Encyclopedia of Integer Sequences.\n\nSquare, Add 2, Subtract 1, repeat...
633940
just divide both sides by 0.015...\n\nand you will get...\n\n566.66 \n\nwith a repeating six.
528120
For 22 * 11 add 2+2=4 and put it in between the orginal number.\n2 4 2\ni.e. 44 * 11 = 4+4=8 = 4 8 4\n\nFor 55 * 11 add 5+5=10 and put it in between the orginal number.\nFor this one, the tens place carrys over to the first digit.\n5+1 0 5 = 6 0 5\ni.e. 66*11 = 6+6=12 = 6+1 2 6 = 726\n\nAlso for 113 * 11 add the first 2 digits (11) + the last 2 digits (13) which 11+13 = 24 and put it between the first and last digit of the three digit number. 1 24 3\ni.e. 126 * 11   12 + 26 = 38 = 1 38 6\ni.e. 149 * 11   14 + 49 = 63 = 1 63 9\ni.e. 314 * 11   31 + 14 = 45 = 3 45 4\nFor this one, the hundred place carrys over to the first digit.\ni.e. 874 * 11   87 + 74 = 161 = 8+1 61 4 = 9 61 4\n\nAlso for 1126 * 11 add the first 2 digits (11) + the middle 2 digits (12) which 11+12 = 23 add the middle 2 digits (12) + the last 2 digits (26) which 12+26=38 and you can see 23 38 but omit the repeat digit in the middle (238) and put it between the first and last digit of the four digit number. 1 238 6\ni.e. 1342*11 = 13+34=47 and 34+42=76  47 76 = 476 = 1 476 2\ni.e. 1451*11 = 14+45=59 and 45+51=96  59 96 = 596 = 1 596 1\n\nThese get a little tricky but its easier than doing it the long way\ni.e. 3884*11 = 38+84=126 and 88+84=172 so you get\n\n3 _ _ _ 4\n1 2 6   \n_ 1 7 2\nHere you choose the larger number for the thousand and hundreds place and forget about the other but add the ten thousands place.\n4 2 7 2 4\n\n4992*11 = 49+92 = 148 and 99+92 = 191\n\n4 _ _ _ 2\n1 4 8\n_ 1 9 1\n5 4 9 1 2
18713
Looking at...\n\n>>>>>>>>> 0\n>>>>>>>>> 1\n>>>>>>>>> 2\n>>>>>> etc.\n>>>> 999999\n\n...in the ones or 10^0 column you repeat\n\n>>>> 0+1+2+3+4+5+6+7+8+9 = 45\n\n100,000 times.  That gives you 4,500,000.\n\nIn the tens or 10^1 column you repeat\n\n>>>> 10*0 + 10*1 + 10*2 + ... + 10*9 = 10*45\n\n10,000 times.  (i.e., group ten 0's in a row, then ten 1's in a row, ..., ten 9's in a row, and repeat that 10,000 times.)  That also gives you 4,500,000.\n\nLikewise with the 10^2, 10^3, 10^4, and 10^5 columns.\n\nSo far that totals 6 times 4,500,000 = 27,000,000.\n\nAll that remains is to add in the 1 from 1,000,000 to get the final total of 27,000,001.\n\nCheck: compare to previous reply. :-)\n\n(How do you get this thing to indent a line?)
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?
508517
SOLUTION WITH WORKING\n----------------------------------------\nSince it is a 2 digit number,say xy i.e.(10x+y in mathematical terms) ,where x is tens' place and y is units'\n\nThe given information says:\n  \n6(x+y) = (10*x)+y\nSo,\n\n6x+6y=10x+y\n4x=5y\n\nIn the given region(10 to 99),searching integer values of x and y,\nwe gwt\nx=5 and y=4 \n\n(Next solution comes when x=10,y=8  but x is single digit number hence this and all further possibilities get discarded.)\n\nHENCE the number is xy=54
219512
Without resorting to a search of the web I have memorized the following digits. (I hope I don't mistype any):\n\n3.1415926535897932384626433832\n795028841971693993751058209749\n445923078164062862089986280348\n253421170679821480865132823066\n470938446095505822317253594081\n284811174502841027019385211055\n5964462294895493038196
463769
lets say that the number is z\n\nthen 10x + y = z.....(1)\nwhere x is the tens digit and y is the units digit.\n\nx + y = 12................(2)\nz = 11x....................(3)\n\nSubstitute (3) into (1)\n\n10x + y = 11x\n-->\ny = x..(4)\nsub (4) into (3)\n\n2y = 12\ny = 6, x = 6\n\nthe number i.e. z = 66