ok: 1 3 5 7 9 11\n    2 4 6 8 10 12\n    3 5 7 9 11 13\n    6 8 10 12 14\n\n\nwhat is a possibility for the next row of numbers???
213932
aren't u missing a number there on the last row???\n\nin that case... i would say...\n\n9 11 13 15 17 19....
541278
Take each number, multiply by two, and then subract two.\n\n3*2 - 2 = 4\n4*2 - 2 = 6\n6*2 - 2 = 10\n10*2 - 2 = 18\n\nedit: I'm sure that the two possible answers are related.  I would also get 34, then 66, then 130 as the next numbers in the sequence.  It has to do with the fact that it started with three.  If we had started at 2, then my system would always yield 2 every time, but you guys wouldn't.  I think both are acceptable, although I will admit that the other method is more interesting.
453627
It's pretty obvious if you write out the multiplication:\n\n......1111\n....x1111\n-----------\n......1111\n....1111\n..1111\n1111\n----------\n1234321\n\nAlgebraically 1111 x 1111 = 1111 x 1 + 1111 x 10 + 1111 x 100 + 1111 x 1000 = 1 x 1 + (1+1) x 10 + (1+1+1) x 100 ....\n\nEach row is 1111 and each row shifts one to the left.\n\nThe nice symmetry would break down after 111111111, because then you'd have to start carrying.
281068
b)two, but only if you are looking for specific Numbers to complete the table.\nThe two would have to be in different row\nYou have 6 blank spaces on the table. \nIf you have one more number given you can find the other column by subtraction.\nIf you have two more numbers given you can find the other row and Column by subtraction.\n\nI can complete the table right now you take 36/64=.5625\nso I use the equations  \n.5625C2+C2=53 1.5625C2=53 53/1.5625=C2=33.92\n.5625C2+C2=26 1.5625C2=26 26/1.5625=C2=16.64\n.5625C2+C2=21 1.5625C2=21 21/1.5625=C2=13.44\nC1=53-C2=19.08\nC1=26-C2=9.36\nC1=21-C2=7.56
516885
C is at the end, so its just the other 8 positions to play with.\n\nCount how many permutations of those 8 there are.\n\nThen count how many there are where A and B stand side by side.\n\nThen subtract those out.\n\nEdit:  Poster below me is way off.  without any restrictions, 9 men could be arranged in over 360000 ways.   Not 81.\n\nBasically, you can take 'C' out of it.  There are 8 spots left, and 8 men.  Spot 1 can hold any of the 8.  spot 2 any of the remaining 7 and so on, giving 8X7X6X5X4X3X2 which is 8! possibilities.\n\nThat's 40320.\n\nOf those 40320, some involve A and B being side by side.  How many?  Well, they could be in slot 1&2, 2& 3 .... up to 7&8.  And they could be A-B or B-A.  That's 14 possibles.  For each of those 14, the other 6 spots could be filled 6! ways.\n\nSo there are 14 X 6! = 14 X 720 = 10800 combos with C in the 9 spot and A & B side by side.\n\nTherefor, of the 40320 with C in the 9 spot, 40320-10800 = 30240 possible arrangements have C at the end, but A and B NOT side by side.
311059
1 - the initial 1\n11 - all numbers from phrase "previous has 1 digit 1"\n21 - all numbers from phrase "previous has 2 digit 1"\n1211 - all numbers from phrase "previous has 1 digit 2, then 1 digit 1"\n111221 - all numbers from phrase "previous has 1 digit 1, then 1 digit 2, then 2 digits 1"\n312211 - all numbers from phrase "previous has 3 digits 1, then 2 digits 2, then 1 digit 1"\n13112221 - all numbers from phrase "previous has 1 digit 3, then 1 digit 1, then 2 digits 2, then 2 digits 1"\n\nSo, the next will be\n\n1113213211 - all numbers from phrase "previous has 1 digit 1, then 1 digit 3, then 2 digits 1, then 3 digits 2, then 1 digit 1"\n31131211131221\n13211311123113112211\n11131221133112132113212221\nand so on
16439
The next line is 312211. First line : 1\nSecond line describes the first line : One-one (11)\nThird line describes the second line : Two-ones(21)'\nFourth line describes the third line : One-two, One-one (1211).\n\nEtc.\n\nSo, the sixth line would be 3 one's, two two's, and one one = 312211
40076
This is kinda wierd without drawing it out, but I'll try to explain and you can come up with the picture.\n\n1)  Start Program\n2)  Accept Input\n3)  Add 1 to counter\n4)  If counter is not equal to 3; go to #2\n5)  If N2 > N3, do #6-8; if not, go to #9\n6)  Move N3 to a hold field\n7)  Move N2 to N3\n8)  Move the hold field to N2\n9)  if N1 > N2, do #6-8 with the N1 & N2 replacing each other\n10) if N2 > N3, do #6-8\n11) print them
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...
501804
560\n2*280\n4*140\n5*112\n7*80\n8*70\n10*56\n16*35\n20*28\n\n300\n2*150\n3*100\n4*75\n5*60\n6*50\n10*30\n12*25\n\n\n222\n2*111\n3*74\n6*37
267908
17\n1, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 21,31,41,51,61.
543635
OOOOOOOOOh i LUV LUVLUVLUV challenge 24!!!! ok her are some ideas:\n7,2,2,1=(7-1)(2+2)\n6,8,2,1=(8/4)(6*1)              ***All taken from Challenge 24 cards\n3,3,6,2=(3+3)(6-2)                     (single digit)\n9,2,5,1=(9-1)(5-2)\n~* ah what the heck i dont have time to finish this. I'llget back to you later, soon. and for future reference, DO buy challenge 24 cardsand so you can use them often. the only annoying thing is, it doesnt come w/anser key u have to figure it out yourself:(
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.
100235
1. Pick a number between 1 and 2023 \n   2\n2. Pick a number between 5 and 1338   \n   613\n3. Pick a number between 1 and 2108  \n   18\nThis is a religious forum :)
505656
26.  The pattern has to start 1, 2, 4, 8, 15.  I see the differences are 1, 2, 4, 7, and the differences of the differences are 1, 2, 3.  The next difference of the difference has to be 4, so, working my way back up, I figure the next difference has to be 11, making the next number 26.
544808
8,6.12,18,10,1,4,21,8,12,7,15
518118
Try to factorise number 319.\nIt is 11 * 29\nSo you have 11 rows 29 people per row \nor 29 rows with 11 people per row.\n\nIf you wonder how to find it it is simple\nTry to divide 319 with prime numbers\n\n2,3,5,7,11,13,17\n\nOnly 11 is ok!