what are the next 2 terms of this series ... 1,11,21,1211,11... ? describe why won't digit 4 come in answer?
463413
1, 11, 21, 1211, 111221, 312211, 13112221, ....\n\nthe odd terms (reading from left to right) enumerate the number of times the same figure (which will be specified to its right) appears in the last term of the sequence.\nSo there can be at most three consecutive equal figures in a term: this happens when a figure appears a number of times equal to itself in the previous term, and also the consecutive figure (different from it) appears the same number of times. As the next figure must be different, it cannot give the same figure as a fourth after the first three.\nFor the same reason, the figure before a sequence of three equal must denote a figure different from the one which has been repeated three times, otherwise the count of that figure would have been different.
986330
It's multiple choice, so all you really need to do is start plugging numbers in and see if they work:\ng(7) = 2(343) -11(49) + 8(7) + 21\ng(7) = 686 - 539 + 56 + 21 =/= 0\n\ng(1) = 2(1) - 11(1) + 8(1) + 21\ng(1) = 2 - 11 + 8 + 21 =/= 0\n\ng(7/2) = 2(343/8) - 11(49/4) + 8(7/2) + 21\ng(7/2) = 343/4 - 539/4 + 112/4 + 84/4 = 0\n\nThus, d is the right answer.
696335
You're buying digits.\nEach digit cost 5 cents.\n1 is one digit.\n21 consists of 2 digits.\n347 consists of 3 digits.
748360
Regroup and add.  You got the 5n, so you should be able to get the other part!\n\nn + n - 1 + n - 2 + n + 7 + n - 11 = \nn + n + n + n + n - 1 - 2 + 7 - 11 =\n5n - 7\n\nNotice that -1 -2 - 11 = -14, add 7 to it and you get -7\n\nNRT
798974
The answer is:\n11=6, 12=6\n\nWhy?\nOne=3 letters\nTwo=3 letters\nThree=5 letters\nFour=4 letters\netc.\n\nI think you can see the pattern now...\nEleven=6 letters\nTwelve=6 letters
967975
2A11\n4D13\n12G17\n48J25\n240M41
1110740
I -1 digit\nLOVE - 4 digits\nYOU - 3 digits
484936
I know for 11x you just have to sum two last digits and insert this sum between the last digits. Ex:\n11x11=121 (we add 1+1 and insert it between 1 and 1)\n11x13=143 (3+1 and insert it between 1 and 3)
830550
143 =13* 11\n122 = 61 *2\n360 = 2 *2*2*5*3*3\n36 = 3*3*2*2\n26 = 13 *2\n234 = 2*3*3*13\n210 = 7*3*2*5\n200= 2*2*2*5*5.....
936927
There is no last digit.  It's a repeating decimal representation of a rational number.  1/3 is such a number, represented by 0.33333...\nThis is the sum of an infinite series, 3/10 + 3/100 + 3/1000 ...  Your answer doesn't have an 8 because that's not the value of the number you cited.  Your series is:\n\n123456790/10^9 + 123456790/10^18 + 123456790/10^27 ...\n\nYour assignment now is to find the rational number represented by the series you want to have:\n1234567890/10^10 + 1234567890/10^20 + 1234567890/10^30 ...\n\nDo your experimenting with smaller series before you tackle the big one.
594189
34,\n\nYou are having the series as 2 times <prime numbers>\n2*2, 2*3, 2*5, 2*7, 2*11, 2*13, 2*17, 2*19, 2*23...\n = \n4, 6, 10, 14, 22, 26, 34, 38, 46 ...
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 :))
1145815
6\n\nYou are putting the single digit numbers in alphebetical order:\neight, five, four, nine, one, seven, six, three, two, zero
887323
1 4 2 5 3\n\nBut it's cyclical, so the sequence is 1 4 2 5 3 1 4 2 5 3 1 4 2 5 3 ... and so on.\nYou can start anywhere in the sequence. The first answer you received is correct, but that person chose to start with 3 instead of 1.
789066
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.
368292
Easy...\n\n312211\n\nLook at it this way...\n\nOne\n1 one\n2 ones\n1 two 1 one\n1 one 1 two 2 ones\n\nTherefor 3 ones 2 twos 1 one
941068
Its a bad question, but the answer they're probably looking for is 8.\n\nIts not prime numbers as some have suggested, because 1 isn't a prime number. (Prime numbers are  2,3,5,7,11 etc)\n\nIt looks like its supposed to be the Fibonacci series, but they've missed off the first 1\n\nThe Fibonacci series is    1, 1, 2, 3, 5, 8, 13, 21 etc