What is the next number in the follwoing sequence? 3,9,27,81,?,?
107194
243\n\nKeep on multiplying by 3.\n3 x 3 = 9\n9 x 3 = 27\n27 x 3 = 81\n81 x 3 = 243\n\nGood luck and have a nice day!
853291
4 is the answer. Here's how:\n4 appears after a break of two numbers, so the number that follows 40 must be 4
676633
54\n*45\n____\n\n 270\n\n2160\n_____\n\n2430
130493
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
688302
453,131,254,369,235,645,641\n\n....and probably more!
526341
You see: \n9\n19\n29\n39\n49\n59\n69\n79\n89\n90\n91\n92\n93\n94\n95\n96\n97\n98\n99\n\n\n\nThat counts 20 9s.\n\nI did not see other answers. When I saw your question, I smiled because I know that you did mislead us as 9, 19... for avoiding the nervous nintees.\n\n\n\nThank you for this nice Question.\n\n\n--- LeeeN
940980
So is this sequence based purely on a mathematical function?\n\nOkay, so there's a 52 in there, could this sequence somehow relate to the calendar?\n\nWhat sort of scientist? I'm a computer scientist and nothing's glaringly obvious about those numbers.\n\n** GOT IT *** 68! Er or Erbium, the sequence spells out Brainteaser based on the atomic weights of B, Ra, In, Te, As, and Er.
572436
Arithmetic sequence:  1, 2, 3, 4, 5, 6, 7\nGeometric sequence:  1, 2, 4, 8, 16, 32, 64
869678
1 and 1\n9 and 3\n25 and 5\n49 and 7\n81 and 9\n121 and 11\n...
471834
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....
1164476
The next number is 3. \nGoing by the number sequence and the number of letters per word in your question then that would be the next number. \nThe final number would be 3 as well.
1121383
36\n95\n1080\n1080\n1080
1157976
0000, 0001, 0002, 0003, 0004, 0005, 0006, 0007, 0008, 0009, 0010, 0011... etc, etc  untill u get to 9998, 9999\n\ntry it out, might take a long time but one combination will work :))
100604
anything like, n.0... , n.1..., n.2..., n.3..., n.4... will be $n.\n\nanything like, n.5... , n.6..., n.7..., n.8..., n.9... will be $n+1.
448118
Oh jeeze, are you serious LOL.......ok lets say: 999,349,843,928,999,344,576,345,768,324,999 gazilliontrillionbazillion........
1158100
adding the number next in sequence. first added 1, then 2, then 3.....
824405
73\n\nThe next one I think is 793