One number is 12 more than another.  The sum of the smaller number and twice the larger number is 39.?
538718
x is the smaller number\ny is the larger number\n\none number is 12 more than the other\ny = x + 12\n\nThe sum of the smaller number and twice the larger number is 39.\n\nx + 2y = 39\n\nfrom here, pick a good solving method substitution or elimination (or even matrices if you really like pain).  I'd recommend elimination:\n\nx + 12 = y      (-)\nx + 2y = 39 \n___________\n0-2y+12=y-39\n\nSolve that for Y.\n\nplus that Y into one of the equations and solve for X.
709150
"Twice" --> 2 *\n"sum of a number and 3" -> (n + 3)\n"is the same as" --> = \n"11 increased by the number" --> 11 + n\n2(n + 3) = 11 + n\n\nDistribute the 2 through the parentheses:\n2n + 6 = n + 11\n\nGet like terms together (subtract n and subtract 6):\n2n - n = 11 - 6\n\nSimplify:\nn = 5\n\nChecking:\nThe sum of 5 and 3 is 8.\nTwice this is 16.\n\n11 increased by 5 is 16, which is the same.\n\nSo the number is 5.
536743
The sum of two numbers is 30\nx+y = 30\n\nThier difference is 6\nx-y = 6\n\nEliminate Y through addition (it's practically eliminated itself)\n\nx+y = 30   (+)\nx-y = 6\n_________\n2x = 36\n\nx = 18\n\nnow \nx + y = 30 right?\n18 +y = 30 (x = 18 so 18 replaces x)\n-18    -18\n___________\ny = 12
695947
D\n\nEither is sufficient.  \n1) will lead to 2 equations,2 unknowns which yield a unique solution for x and y:  x=y+1, x+y=77, therefore x=39, y=38.  xy=1482 \n\n2) If the sum is 77 and x and y have the same tens digit, then there are two solutions for {x,y}, namely {38,39} and {39,38}, however the product xy is the same. xy=1482
831094
36. >\n38. =\n40. >\n42. <\n44. <
1123620
<  Less Than\n> Greater Than
264056
k=12, assuming you meant 4(k+2) - 2(k-1) = 30, otherwise it won't work out.
632560
There is no simple trick for the addition of two numbers. You will have to calcuate their actual value and then add them. So for you example you would have to go to 4 + 8 = 12.
1093015
41\n\n11+2=13\n13+4=17\n17+6=23\n23+8=31\n31+10=41\n41+12=53\n53+14=67
942916
your question has a problem...............\nif the sum is 13, then the other number has to be -35, which will not meet your other condition
748322
The numbers for this particular answer are 9 and 10.\n\n9 + 10*3\n9 + 30\n=39
429784
144/12 = 12*13 = 156 + 32 = 188
93238
You must be very confused. \n\n1+1 = 2 plus one more is 3
320053
Phew, hard question.\n\nSTEP 1: minus the remainders from the numbers. you would get 2923 and 3239 respectively.\n\nSTEP 2: this is the hard part. take both the numbers individually and take their factors.\n\nfor 2923 you wold get 37*79, and for 3239 you would get 41*79. you get this by painstaikingly checking with each prime number whether or not the number is divisible by it or not.\n\n37 is the 12th prime number (excluding "1") and 41 is the 13th (again excluding "1"). This means you have to do 12+13=25 calculations. then divide the main number you got from STEP 1 with 37 and 41 respectively, and you will get 79 in both case.\n\nSTEP 3: Notice that 79 is common in both the cases. This is the greatest common number.\n\nIf there were more than one common number, you would just multiply all the common numbers to get the answer.\n\nHAVE FUN
825140
eleven million four hundred thirty one thousand = 11431000.\nfive million five hundred forty four thousand = 5544000.\nSo the difference is 5887000.
1103226
N = the number\nN - 2   <<<   2 subtracted from the number\n( N - 2 ) x 3   <<<   the difference tripled\nN + 6   <<<   6 more than the number\n\n( N - 2 ) x 3 = N + 6\n3N - 6 = N + 6\n2N = 12\nN = 6
526416
add the numbers\n15\n19\n42\n19\nnext number equals 15.....     and each has a 2 and a 7 in it\n\ntry 2742 sounds like a good answer for neopets lenny conundrum answer possibility but not fully sure..... lucidlyinsane