I need to find two consecutive multiples of eight whose sum is 184...?
963648
Well, the people above me did all the work. But your correct answer is 88 and 96.
320882
1 2 3 4 5 6 7 8\n<---[======)--->
748660
The answer is 78:\n\n1² + 2² + 3² + 8² = 1 + 4 + 9 + 64 = 78\n2² + 3² + 4² + 7² = 4 + 9 + 16 + 49 = 78\n1² + 4² + 5² + 6² = 1 + 16 + 25 + 36 = 78\n\nI did this creating a spreadsheet with the first 9 squares.  There were 126 ways (9 choose 4) ways to pick the four squares which then resulted in a sum.  I sorted the sums and found the one that repeated itself 3 times (78).\n\nThe answer is 78.
166778
29786\n+  850\n+  850\n_____\n31486\n\nThe way to solve this is to first note that the units digit of Y + N + N is Y. Therefore N must be 5 or 0. If N is 5 then you'd need to carry the one so that now the units digit of 1 + T + E + E is T. But this is impossible because E + E is an even number. Therefore N is 0. Using similar logic, we find E must be 5 and we do carry a one to the next column. Let's look at what we have now:\n\n___1\nFORTY\n+__T50\n+__T50\n_____\nSIXTY\n\nNow, recognize that T50 and RTY are less than 1000 and FO thousand plus two thousand (and change) can only equal SI thousand if O is 9 and I is 1. This gives:\n\n___1\nF9RTY\n+_T50\n+_T50\n_____\nS1XTY\n\nF and S must be consecutive and R and X are not both odd or both even (since we carried the one). If R is 8,=> T is 7,=> X is 3, but there remain no consecutive numbers available for F and S, so R is not 8, yet 1 + R + T + T must be at least 22, so T is therefore 8. R = 6 also fails to allow consecutive values for F and S, so R = 7.\n\n___1\nF978Y\n+_850\n+_850\n_____\nS1X8Y\n\nFrom here it should be easy to see X is 4 and F and S are 2 and 3 respectively leaving Y to be 6 and solving the puzzle.
95089
1(1)+0(2)+1(4)+1(8)+1(16)+0(32)+0(64)+0(128)+1(256)=\n1+0+4+8+16+0+0+0+256=\n285
845925
If you problem, or part of it, is:\n\nHow many 8 digit numbers can be formed from this list of digits?\n8! / (3! * 2!)\nThe number combinations of digits has tobe reduced by 2! because of the 2 "6"s.\nThe number combinations of digits has tobe reduced by 3! because of the 3 "2"s\n.\nBut since the number cannot exceed 65,500; all larger numbers must be eliminated. Subtract the number of 8 digit numbers that can be produced from the list that start with "9". Likewise "7". Likewise "69", "67", "66", "659", ""657", and "656".\n\n\n\n(Try to state your problem clearly and completely.)
1025141
The series is very simple.\n\n0  1  1  2  3  5  8  13  21  34  55  89  144  ... \n\nYou just add the two previous numbers to the next number.\n\nThere are tons of websites related to your question, so you just haven't looked, and so don't know enough to ask a specific question.
975052
888 + 88 + 8 + 8 + 8 = 1000\nThat's eight eights.
820779
Consecutive integers are n, n + 1, n + 2, n + 3\n\nThe sum of the second and fourth equals the third increased by 17.\n\n(n + 1) + (n + 3) = (n + 2) + 17
9961
0,1,1,2,3\n\nto get the next number add the previous 2 numbers
394510
Three Six Nine Twelve Fifteen Eighteen Twenty-one Twenty-four Twenty-seven....multiples of 3.
861916
I can't see the rest of your question, but even numbers can be multiples of three. Like 6 (2x3). They are not mutually exclusive.
53832
First you try to get all of the parts to have same denominator (which looks like 8 is what you need).\nSince the first and the last are already w/ 8, then u need to * and / the middle number with 2.\nSo you get:\n2/8-4/8+5/8\nnext you can write:\n(2-4+5)/8\nand that's 3/8\n=>\n2/8-2/4+5/8 = 2/8-4/8+5/8 = (2-4+5)/8 = 3/8
376218
Not sure how many you need, but here are the first few:\n2,3,5,7,11,13,17,19,23,29,31, 37,41,43,47,53,59,61,67, 71,73,79,83,89,97,101
362147
8 X 11=88\n\n88 X 154 = 13552\n\n13552 +76= 13628\n\n13628 -3 = 13625\n\n13625 + 7= 13632\n\nTHERE THERE ALL BETTER
1166207
2[w+ {2w+8}]= 64\n3w+8 = 32\nw= 8 \nyour are right
252544
interesting!!!!  \n\n                 9x8 =72  7+2 =9\n                 9x9 =81  8+1 =9\n                 9x3 =27  2=7 =9