i found something for my self,and i want that someone can explain me why is like that?\n\n                 1 2 3 4 5 6 7 8 9 \n                 2 4 6 8 1 3 5 7 9\n                 3 6 9 3 6 9 3 6 9 \n                 4 8 3 7 2 6 1 5 9 \n                 5 1 6 2 7 3 8 4 9 \n                 6 3 9 6 3 9 6 3 9 \n                 7 5 3 1 8 6 4 2 9\n                 8 7 6 5 4 3 2 1 9\n                 9 9 9 9 9 9 9 9 9\n\ni found in that way: for ex:  3*3=9  \n                              4*4=16=1+6=7\n                              8*8=64=6+4=10=1+0=1\nwhen you take out the single numbers all the numbers are build from nine? 9? we can see that here:\n   1 0 0 0 0 0 0 0 0     0 0 0 0 0 0 0 8 0\n   0 0 0 0 1 0 0 0 0     0 0 0 8 0 0 0 0 0\n   0 0 0 0 0 0 0 0 0     0 0 0 0 0 0 0 0 0\n   0 0 0 0 0 0 1 0 0     0 8 0 0 0 0 0 0 0\n   0 1 0 0 0 0 0 0 0     0 0 0 0 0 0 8 0 0\n   0 0 0 0 0 0 0 0 0     0 0 0 0 0 0 0 0 0\n   0 0 0 1 0 0 0 0 0     0 0 0 0 8 0 0 0 0\n   0 0 0 0 0 0 0 1 0     8 0 0 0 0 0 0 0 0 \nwhy is strange?
129113
You haven't explained the question properly! These matrix can be formed any way!
603855
the number is  : 64665912\n\nnow why? i ll show you\n\nfirst off all i denote the number by 8 different symbols\n\nq w e r t y u i \n\nq and r are obviously 6 \n\nso if you write down your sums you have\n\n6 t y u i \n+  6 w e \n------------\n6 6 5 5 8\n\nso y is 9 \n     t is 5\n\nnow go to the other sum\n\n6 w e 6 5\n+    9 u i \n-----------\n6 5 5 7 7\n\nso e = 6 \n     w = 4\n     i = 2\n     u = 1\n\nso you have your number 64665912\n\n\n\nP.S. Thanx Cosmin C below for the explaining of the obviously part!
6083
Oddities like this are due to floating-point math, and are quite common, unfortunately.  The problem is that floating point numbers are stored in binary, and so they can only be an approximation of most numbers (like .1).  I bet if you do this calculation in Python or C, you'll get the same sort of answer.
514903
Hello,\n\nThe symbols in relationship to the numbers change each time. \nThe symbol you choose is the correct one  for all answers. No matter the number you choose there is one correct symbol for all answers.
317366
try seeing OEIS: http://www.research.att.com/~njas/sequences/?q=1%2C+1%2C+3%2C+5%2C+11%2C+21%2C+43&sort=0&fmt=0&language=english&go=Search\n\neverytime you had an integer sequence problem, you could check there, most of the time, they had an entry of the possible sequence there.\n\nthe most possible answer to your question is perhaps the \nJacobsthal sequence: a(n) = a(n-1) + 2a(n-2). \n\nwhich gives the sequence\n0, 1, 1, 3, 5, 11, 21, 43, 85, 171, 341, 683, 1365, 2731, 5461, 10923, 21845, 43691, 87381, 174763, 349525, 699051, 1398101, 2796203, 5592405, 11184811, 22369621, 44739243, 89478485, 178956971, 357913941, 715827883, 1431655765, 2863311531, ....
281955
1) The sum of the positive 3-digit odd integers = 101 + 103 + 105 + ... + 995 + 997 + 999.  Adding the first of these to the last, we get 101 + 999 = 1100.  Adding the second of these to the next to last, we get 103 + 997 = 1100.  This pattern keeps repeating until 549 + 551 = 1100.  The question then becomes, how many of these 1100's do we have?  We have one for 101, 103, 105, ..., 549 (the later numbers share their 1100's with these).  How many numbers are in the set 101, 103, 105, ..., 549?  Well, there are 449 numbers in 101, 102, 103, ..., 549.  Since there is one more odd number than even number, we must have 225 odd numbers and 224 even numbers.  Thus, there are 225 odd numbers in the set 101, 103, 105, ..., 549.  Thus, we have 225 1100's to add together.\n\n225 * 1100 = 247500.\n\n\n\n2) This time we wish to add the numbers of the set 10, 11, 13, 14, ..., 94, 95, 97, 98.  Adding the first to the last we get 10 + 98 = 108.  Adding the second to the second to last we get 11 + 97 = 108.  The pattern keeps repeating until we get 53 + 55 = 108.  How many 108's do we have?  Well, we have one for 10, 11, 13, 14, 16, 17, ..., 52, 53 (again, the later numbers share their 108's with these).  How many numbers are in this set?  In the set 10, 11, 12, 13, ..., 52, 53, there are 44 numbers.  The multiples of three range from 12 to 51--from 3*4 to 3*17.  Thus, there are 14 multiples of three in this range.  This means that there are 30 numbers in the set 10, 11, 13, 14, 16, 17, ..., 52, 53.  Thus, we have 30 108's to add together.\n\n30 * 108 = 3240.\n\n\n\n3) Here, I am assuming that the raise of 10% also applies to the previous year's raise and not just to the "principal" of $18000.  Thus, I am assuming a compound raise instead of 5 simple raises.\n\nDuring the first year, you make $18000.  During the second year, you make 18000 * 1.1 = $19800.  During the third year, you make 19800 * 1.1 = $21780.  During the fourth year, you make 21780 * 1.1 = $23958.  Finally, during the fifth year, you make 23958 * 1.1 = $26353.80.\n\nGood luck.
214487
If you look at the symbol to the right of the number 9 on the chart, and then the one next to the 18, and 27, infact, the symbols next to all multiples of 9 are the same.\n\nIf you subtract the quantity of each digit from any two digit number, you will always end up with a multiple of 9. So, there you go. It's known ahead of time that your nunmber will be a multipe of 9 so all multiples of 9 have the same symbol and that's the one that appears in the magic box.
600709
every time i've asked a teacher this, they've told me that its an even number.  The only reason i can come up with is that it is found between  two odd numbers. (-1 and 1)
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
484475
ok look if the #, 538 last # is even then it is divisible by 2. And the rule for 3 is you add up all the digits 5+3+8=16, and if that #, 16, is divisible by 3 then the whole # is divisible by3. The rule for 5 is if the #, 538, ends in a 0 or 5. The rule for 6 is add up all the digits again 5+3+8= 16 and if 16 is divisible by 6 then the whole # is divisible by 6. And the rule for 10 is if the # ends in a 0!! \ngood luck
408290
Firstly, it has to start with either a 1 or a 2, and end with a 2 or a 6. So we just take each case one at a time.\n1..2: that leaves 23556, so there are 5!/2! ways, ie 60 ways.\n1..6: that leaves 22355, so there are 5!/(2!2!) ways, ie 30 ways.\n2..2: that leaves 13556, so there are 5!/2! ways, ie 60 ways.\n2..6: that leaves 12355, so there are 5!/2! ways, ie 60 ways.\n\nNote that in each case there are 5! ways of arranging the other 5 numbers, but we need to divide by 2 if there is a pair of repeated numbers.\nThat gives a total of 210.\n\n(note that both of the other answers given so far are wrong - its not 360, because that person has forgotten they have to be even; and its not 120. They counted 2x2 possibilities for the first and last digits which is correct, but one of those cases already uses both 2s, so you *don't* want to divide by 2 in the end for that particular case (you're not counting them twice are you!)\nTheres really no other way of doing it than splitting them into cases like I did.
235482
yeah, looks like the circle is just a 0, which would mean x^0 + 7 = 1 + 7 = 8.  It does look weird, though.
494264
Here is a super quick guide to learning to write numbers in binary...\n\nbinary uses 1/0 (on and off) to give value to numbers. This is why it is best used in computer language, because the computer can understand on and off. Now look at the following...\n\n1     \n2      \n4       \n8        \n16    \n32    \n64      \n128   \n\nNow,... notice the above... each number is equil to the number above it x2. this is the basis in learning binary...\n\nlets say we want to write the number  37....  we do the following...(read it from the bottom up)\n\n\n1        (there is one 1 in 1, this leaves us with 0 left over...)\n2        0\n4        (there is one 4 in 5, this leaves us with 1 left over....)\n8        0\n16      0\n32      (there is one 32 in 37, this leaves us with 5 left over...)\n64      0\n128    0\n\n\nso,... in binary, 37 would be   101001  (one 1, one 4, one32)\n\n\nusing the above we can learn to count to 10 !!!\n\n1   (1=1)\n01  (2=2)\n11  (2+1=3)\n001 (4=4)\n101 (4+1=5)\n011 (4+2=6)\n111 (4+2+1=7)\n0001 (8=8)\n1001 (8+1=9)\n0101 (8+2=10)\n\n\n\nthis is a very simple view of binary...\n\n\nhope this helps!!\n\n\n=)
640634
N = 285714 and 1.5*N = 428571\n\nHere's the analysis:\n\nLet's assume that it's an n digit number with digits x_1, x_2, ...  x_n. Then\n  N = Sum{i=1..n}(10^(n-i)*x_i)\nand the number you get by moving the last digit to the front will be:\n  10^(n-1)*x_n + Sum{i=1..n-1}(10^(n-i-1)*x_i)\nWhat you are asking for is:\n  1.5*Sum{i=1..n}(10^(n-i)*x_i) \n    = 10^(n-1)*x_n + Sum{i=1..n-1}(10^(n-i-1)*x_i)\nor:\n  1.5*x_n + 1.5*Sum{i=1..n-1}(10^(n-i)*x_i) \n    = 10^(n-1)*x_n + Sum{i=1..n-1}(10^(n-i-1)*x_i)\nor by transposing to the other side:\n  1.5*Sum{i=1..n-1}(10^(n-i)*x_i) - Sum{i=1..n-1}(10^(n-i-1)*x_i)\n    = 10^(n-1)*x_n - 1.5*x_n\nor: \n  Sum{i=1..n-1}(1.5*10^(n-i)*x_i) - Sum{i=1..n-1}(10^(n-i-1)*x_i)\n    = (10^(n-1)-1.5)*x_n\nor:\n  Sum{i=1..n-1}(15*10^(n-i-1)*x_i) - Sum{i=1..n-1}(10^(n-i-1)*x_i)\n    = (10^(n-1)-1.5)*x_n\nor:\n  Sum{i=1..n-1}(14*10^(n-i-1)*x_i)\n    = (10^(n-1)-1.5)*x_n\nor:\n  14 * Sum{i=1..n-1}(10^(n-i-1)*x_i)\n    = (10^(n-1)-1.5)*x_n\n\nNow the left hand side is a positive integer and on the right hand side 10^(n-1)-1.5 will be a fraction ending in 0.5. So x_n has to be an even digit to make (10^(n-1)-1.5)*x_n a positive integer. \n\nNow, if x_n were 2 or 6, multiplying it with something-point-five will give you an odd number. But the left hand side is an even number (because you are multiplying by 14). So x_n cannot be 2 or 6. That leaves 4 and 8 as the two possible choices for x_n i.e., the number N has to end in either 4 or 8.\n\nIf x_n is 4, then we are looking for:\n  14 * Sum{i=1..n-1}(10^(n-i-1)*x_i)\n    = (10^(n-1)-1.5)*4\nor:\n  14 * Sum{i=1..n-1}(10^(n-i-1)*x_i)\n    = 4*10^(n-1)-6\nor: \n  7 * Sum{i=1..n-1}(10^(n-i-1)*x_i)\n    = 2*10^(n-1)-3\nor:\n  7 * Sum{i=1..n-1}(10^(n-i-1)*x_i)\n    = 2*10^(n-1)-3+10-10\nor:\n  7 * Sum{i=1..n-1}(10^(n-i-1)*x_i)\n    = 2*10^(n-1)-10+7\nor:\n  7 * Sum{i=1..n-1}(10^(n-i-1)*x_i)\n    = 20*10^(n-2)-10 + 7\nor:\n  7 * Sum{i=1..n-1}(10^(n-i-1)*x_i)\n    = 10*(2*10^(n-2)-1) + 7\nor:\n  Sum{i=1..n-1}(10^(n-i-1)*x_i)\n    = 10/7*(2*10^(n-2)-1) + 1\n\nAgain, the left hand side is a positive integer. The right hand side can be a positive integer only if 2*10^(n-2)-1 is a multiple of 7. \n\nFor values of n, 2*10^(n-2)-1 is 1(n=2), 19(n=3), 199(n=4), 1999(n=5), 19999(n=6), etc. n=6 is the first number in this series (19999) that's a multiple of 7. Putting n=6, we get:\n  Sum{i=1..5}(10^(5-i)*x_i)\n    = 10/7*(2*10^(6-2)-1) + 1 = 28571\ni.e., a five digit number = 28571.\n\nThat gives us our first answer: N = 285714 and 1.5N = 428571.\n\nNow I have only done the analysis so far for x_n being 4. If you do similar analysis for x_n=8, you get\n  14 * Sum{i=1..n-1}(10^(n-i-1)*x_i)\n    = (10^(n-1)-1.5)*x_n\n    = (10^(n-1)-1.5)*8\n    = 8*10^(n-1)-12\n    = 8*10^(n-1)+2-14\nor:\n  Sum{i=1..n-1}(10^(n-i-1)*x_i)\n    = (4*10^(n-1)+1)/7 - 1\n\ni.e., 4*10^(n-1)+1 has to be a multiple of 7. For various values of n, 4*10^(n-1)+1 gievs you numbers like 41, 401, 4001, 40001, etc. and the first number in this series that's a multiple of 7 is 400001 for n=6. So we end up with:\n\n  Sum{i=1..5}(10^(5-i)*x_i)\n    = (4*10^(6-1)+1)/7 - 1 = 57142.\n  \nThat gives us our second answer: N = 571428 and 1.5N = 857142.\n\nWhile this is also correct, the first number is smaller. So the smallest number that satisfies your constraint is 285714.
88914
Only a few answers are possible. One of those is 18, which is used in the example.  All of the possible answers have the same symbol as 18.\n\nSorry, I don't know how to prove this, but try a few examples, always comparing the symbol representing your result to the symbol representing 18.\n\nOkay, I figured it out. \n\nState the puzzle another way:\n\nChoose two one-digit numbers, x and y.\n\nCalculate 10x + y - x - y = 9x\n\nThe result of every calculation is merely 9x.\n\nIn the chart, every multiple of 9 is represented by the same symbol, so no matter what numbers you choose, you get the same symbol.  \n\nWhen you click on the "Click here to try again" link, it merely redraws the table with a different symbol representing the multiples of 9.
611345
that's really weird.  The solution of having at least one of the letters as zero appears to work.  However, if you play around with the equation, it seems to give you some impossibilities:\n\n9 = DCBA/ABCD ==> 9 = 1 which is impossible\n\nor\nif you put a zero in the above equation you get 9 = 0/0 which is also impossible.\n\nWeird!!!!
70337
There is no flaw as long as you recognize that .99999 repeats forever.  If you were to end it at some point and subtract, then as a previous poster said, you'd actually get 8.99....1.  But since it repeats, there is always another 9 and so it works.\n\nAnd consider this - if .9999999repeating did NOT equal 1, then there would be another number between them.  What would that number be?\n\n~~~~~~~~~~~~~~~~~~~~\nNo, the number in question is not infinitessimal, it doesn't exist.  You can never get to the end of the string of nines to try and squeeze another number in there :)