Wt is 1001th term of seqence 3,7,11,15.....?
613494
Let n be the 1001th term of this series.\nNow, 7-3=11-7=15-11=4\n\nSo 4 is the common difference(d)\na is the first term=3\nNow the 1001th term is = a+(n-1)d\n                       = 3+(1001-1)4\n                       = 3+4000=4003\n4003 is the 1001th term
83400
11\nt h e a l p h a b e t
49819
Hold on....10....20......25.....29..30....25..29\n\n25
1069170
13212211\n\nJust counting numers in the previous number.\n\n'm i right??
808688
the 1st three terms means n=1, n=2, n=3 so substitute\n\nThis gives, 3*1+2,3*2+2,3*3+2\n\n=5,8,11
291700
1\n2\n3\n4\n5\n6\n7\n8\n9\n10
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 :))
75832
#1, #2, #3, #4.......#18, #19, #20.\n\nVoila!\n\n:o)
729648
007103  \n1.  D  Correct   \n  2.  C  Correct   \n  3.  A  Correct   \n  4.  C  Correct   \n  5.  A  Correct   \n  6.  A  Correct   \n  7.  B  Correct   \n  8.  D  Correct   \n  9.  A  Correct   \n  10.  A  Correct   \n  11.  A  Correct   \n  12.  D  Correct   \n  13.  A  Correct   \n  14.  B  Correct   \n  15.  D  Correct   \n  16.  C  Correct   \n  17.  D  Correct   \n  18.  D  Correct   \n  19.  C  Correct   \n  20.  B  Correct  \n\n007104\n\n1.  B  Correct   \n  2.  B  Correct   \n  3.  A  Correct   \n  4.  A  Correct   \n  5.  C  Correct   \n  6.  C  Correct   \n  7.  B  Correct   \n  8.  A  Correct   \n  9.  D  Correct   \n  10.  A  Correct   \n  11.  A  Correct   \n  12.  C  Correct   \n  13.  B  Correct   \n  14.  A  Correct   \n  15.  D  Correct   \n  16.  D  Correct   \n  17.  A  Correct   \n  18.  A  B  PAGES 24-33  \n  19.  D  Correct   \n  20.  A  Correct
804536
1 + 7 + 7(6) + 42(5) + 210(4) + 840(3) + 2520(2) + 5040(1) = 13,700
131067
last digit of 3^17=3 because 3^1=3,3^2=9,3^3=27,3^4=81.Then the last no.repeats four times.so 17/4=4&remainder 1.therefore 3^1=3 =last number.\nlast digit of  7^13=7 because 7^1=7,7^2=49,7^3=243,7^4=1701.Then the last no.repeats four times.so 13/4=3&remainder 1.therefore 7^1=7 =last number.\nTherfore last number=7+3=10.So last number =0
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.
150167
(a) 111100111101\n(b) 110001011
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
368833
10 - B.\n11 - A.\n12 - C.\n13 - B.\n14 - B.\n15 - B.\n16 - C.\n17 - C.\n18 - A.\n19 - A.\n20 - C.
9961
0,1,1,2,3\n\nto get the next number add the previous 2 numbers
532000
How about 9! 51, 52, 55, 56, 57, 58, 59, 60, 61