consecutive integers?
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
102547
A begining and no ending: Positive Integers\nAn ending and no begining: Negative Integers
560969
23\nif intead of '=' there is +
387072
The primes in the 1st answer aren't consecutive.\n\nHere's a list just  using the prime numbers up to 1,000\n\n149-139\n191-181\n251-241\n293-283\n347-337\n419-409\n431-421 --- two in a row here\n557-547\n587-577\n641-631\n701-691\n719-709\n797-787\n821-811\n839-829\n929-919\n\nAnother interesting note, the highest difference between the primes from 1-1,000 was only 18.  And I think that only came up once.
925404
OK, let's see.\n\nIt's an arithmetic sequence, so the difference between consecutive terms is a constant.  The first term is 1, so we can express each term as:\n\nx(i) = 1 + k*(i-1), i = 1, 2, 3, ....\n\nNext the problem says that the 2nd, 10th, and 34th terms are the first three terms in a geometric sequence.  That means that the ratio of the 10th term to the 2nd term must be the same as the ratio of the 34th term to the 10th, because when you divide any two terms of a geometric sequence, you get the same constant.\n\ni(2) = 1 + 1*k\ni(10) = 1 + 9*k\ni(34) = 1 + 33*k\n\nSo we said i(10)/i(2) must equal i(34)/i(10):\n\n1+9k    1+33k\n-------- = --------\n1+1k      1+9k\n\nCross multiplying: (1+9k)^2 = (1+1k)(1+33k).\n\nThen 1 + 18k + 81k^2 = 1 + 34k + 33k^2.\n\nPutting it all on the left side, 48k^2 - 16k = 0, or 16k(3k - 1) = 0.\n\nSo either k = 0 (not interesting) or k = 1/3.\n\nTherefore the arithmetic sequence must be:\n\n1, 1 1/3, 1 2/3, 2, 2 1/3, 2 2/3, 3, 3 1/3, ....\n\nThe second term is 1 1/3.  The 10th term is 4 (1 + 3 * 1/3).  The 34th term is 12 (1 + 33 * 1/3).  So the geometric progression would be:\n\n4/3, 4, 12, 48, 192, .....
1183323
Mike and Pankaj are both wrong.  The answer is 9.\n\nIf a-b=1, then we are looking at two consecutive integers.\n\nIf a^3-b^3=61, then the consecutive integers must be 4 and 5--those are the two integers where the difference of the cubes is 61.\n\nSo 5^2-4^2 is what we are seeking:  25-16=9
906761
Try this:\n\nvoid addIt(int* begin, int* end, int x)\n{\nint* p;\nfor (p=begin; p!=end; ++p) {\n*p = *p + x;\n}
845089
Ok, your question makes absolutely no sense.  Why say we can only use 1 through 16, and then say we can only use 1?\nSome ways, assuming I translated your question correctly, would include 16 + 16 +1 +1, and 16 + 15 + 3.... There are almost an infinite number of ways to add to 34 with the numbers 1-16, so please rephrase your question.
129538
The next number is 30651.\nIt has a recursion, a_n=a_(n-1)^2+2*a_(n-2).
917997
Simply loop through the list of numbers and compute:\n\n'number' modulo '2'\n\nIn C, that would be:\n\nnumber % 2\n\nA modulus of 2 would return a '1' for odd numbers, and '0' for even numbers (since even numbers divide by 2 with no remainder, and odd numbers divide by 2 with a remainder of 1).\n\nExample:\n\nint index;\nint numArray[25];\n\n...(store the 25 numbers in the array)...\n\nfor(index = 0; index < 25; index++) {\n     if(numArray[i] % 2) {\n          printf("%d&#92;n", numArray[i]);\n     }\n}
914435
x > 5\n\n- - - - - -s-
455350
Kind of, if you look at the numbers below:\n\n1,4,9,16,25,36,49,64,...\n\nYou can see that they are increasing by the numbers:\n\n3,5,7,9,11,13,15,...\n\nSo you can find the next square number if you have two preceding ones
1067700
Let the numbers be n and n+1 (consecutive)\n\nthen the answer is 2n(n+1)\n\nYou know, I can tell you're really trying, (not just fishing for homework answers) because you didnt ask the whole question, just the part you needed help with setting up properly.\nGood Luck!
756753
8) The sum of consecutive integers starting with 1 and ending in n is: s=n*(n+1)/2\nThe average of that is s/n\nAnd the effect of removing the mystery number (call it x) from the sum is:\n(s-x)/n=(n+1)/2-(x/n)=31.25\nAs the first term is a multiple of 1/2 in order for this to be true x must be 1/4 or 3/4 of n (it must be less than n remember?).\nTesting 1/4:\n(n+1)/2-1/4=31.25\nn+1=63\nn=62\n\nAs 62 isn't a factor of 4 and we need n to be of the form 4*x (in order for x/n to be 1/4) we can't use this n.\n\nTesting 3/4:\n(n+1)/2-3/4=31.25\n(n+1)=64\nn=63\n\nAs n isn't a factor of 4 there is no solution (n needs to be of the form 4*x/3).\n9) (2^1999)*(5^2001)= (5^2)*(2^1999)*(5^1999)= 25*10^1999\nAs all those 0's aren't going to affect the sum we can remove them. Result: 25\nThe sum of 2 and 5 is 7\n10) x*x + 6x - 91 = (x + 13)(x - 7)\nThe working out is mostly experimentation I'm afraid, but at least the result works. :)\n11) Sorry, no idea how to solve that one.\n12) I'm not sure I understand the question but if it's what I think it is then I can't solve it through non-experimental means. Try looking up combinatorics on wikipedia.
697965
every number from 0 to 10\n\nplus placed in alphabetical order
121013
if it is in binary, 1 + 1 = 10
780025
x ≤ 5\n\n-- - - - - - -s-