Find two consecutive integers such that the sum of 8 times the first integer and 6 times second intege is 118?
479654
Let first integer be x\n=> second integer will be (x+1)\nfrom question-\n8*x + 6*(x+1)=118\n=>8x + 6x + 6 =118\n=>14x =118-6\n=>14x =112\n=>x =112/14\n=>x= 8\n\nthus, the numbers are x=8, and x+1=9; i.e 8 & 9
935700
two consecutive numbers could be called x, and (x+1). Add these two together to find the sum. x+(x+1) = 85.\nSo 2x+1 = 85.\nsubtract 1 from both sides and we have 2x = 84.\nDivide both sides by two to get x by itself so x = 42.\nSo if x=42, the consexutive numbers from the beginning of the problem would be 42, and 43.
78537
So easy:\nheigt 2 than 36/2 =18\nany multiplication of two number give us 18\nthan 6x3=18\nthan h=2\n     l=6\n     w=3
279409
Find every number between 101 and 999 that is divisible by 7 and then add them together.  Hmmm......like I really am going to do that.
237350
Do you mean permutations?\n8P4\n= 8!/(8-4)!\n=8!/4!\n= 8*7*6*5\n= 1680
316069
The formula for the sum of N consecutive numbers starting with number M is, \nM + (M+1) + (M+2) + ... + (M+N-1)  is:\n\n   (2M+N-1)*N\n   ----------\n       2\nfor 50 consecutive numbers we have N=50 and M=1998\n\nthe sum is Sum=(2*1998+50-1)*50/2=4045*25=101125\n\nfor 25 consecutive numbers N=25, Sum=101125, M=?\n(2M+25-1)*25/2=101125\n(2M+24)=8090\n2M=8090-24=8066\nM=4033\nthe smallest number of the 25 consecutive numbers is 4033
1063322
three consecutive even integers are\nx, x+2 and x+4\n\nEquation is\n2((x+2)+(x+4)) = 6x - 12\n\nSolving for x\nx = 12\nSo, 12, 14 and 16
1008114
This series is n(n+1)\n\n2,6,12,20,30,42,56,72,90,. . .
252544
interesting!!!!  \n\n                 9x8 =72  7+2 =9\n                 9x9 =81  8+1 =9\n                 9x3 =27  2=7 =9
1030604
3,3,8\n\nThe reason is that if you write down all possible three integers whose product become 72 you reach following sets, whose sum are written also:\n1,1,72 => 74\n1,2,36 => 39\n1,3,24 => 28\n1,4,18 => 23\n1,6,12 => 19\n1,8,9 => 18\n2,2,18 => 22\n2,3,12 => 17\n2,4,9 => 15\n2,6,6 => 14\n3,3,8 => 14\n3,4,6 => 13\n\nThe reason that the second friend was unable firstly to figure out the values is that the number of building was 14 (since all other summations are unique, therefore he then could easily find out) and the additional explanation of the first friend made him sure that there are no twins in elder age (so the oldest was meaningless). Therefore, we can judge that they should be a twin at 3 and the oldest at 8.
1092827
It is just like your other question.\n3*8*6 = 144
1118957
The first number has to be (x+6) and the second number is (x).  There are four possible combinations: 6,0 (60); 7,1 (71); 8,2 (82); 9,3 (93).  60 plus 8 (68) does not equal 06, or 6.  71 plus 8 does not equal 17.  82 plus 8 does not equal 28.  93 plus 8 does not equal 39.\nUnfortunately, this problem is impossible.\nSorry I couldn't be of more help.
136837
888 + 88 + 8 + 8 + 8 = 1000
137681
Ok...\nI see what you are doing...\nYou merely substitute the number in for x...\nso the problem becomes 6-2*(-1)\n6+2\n8
939412
The way to solve this would be to let the 2nd consecutive number be x. So the three even consecutive numbers are x-2, x and x+2.\n\n(x-2)(x)(x+2)=85008\n\nThis equation can be difficult to solve, but a short method would be to just cube root 85008 to get an approximate middle number.\n\nCube root (85008) = 44\n\nTry multiplying 42x44x46 = 85008 (voila!)\n\nThe numbers are 42, 44 and 46.
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.
870635
(2-3i)(4+6i) = 8 + 12i - 12i - 18i^2\n= 8 + 18\n26