find five consecutive numbers whose sum is 150.?
372053
Choose x to represent the first number.\n\nThat means that\n\nx + (x + 1) + (x + 2) + (x + 3) + (x + 4) = 150\n\n5x + 10 = 150\n\nx = 28\n\nThat means 28, 29, 30, 31, 32
584167
7+3+5=15\n6+7+2=15\n8+2+5=15
913215
The third answer (from samoht14) comes closest to answering this question.  From it you can see that you need to find a value of x so that 2*x + 1 is equal to the odd number.\n\nso, given an odd number, subtract 1 and halve the result to get x.\n\nThen x and x+1 are the numbers you want.\n\ne.g. for 37, \n36/2 =18 and so the consecutive integers are\n18 and 19.\n18 + 19 = 37
545435
I would think 34 because I see a pattern at the end\n19 - 5 = 14\n14 + 20 = 34(You cant subtract unless your looking for integers which then would be -6 because:\n14 - 14 = 0 + 6 = 6 and you put the negative sign)
702073
30 30 150 180 1200\n\n100 100 100 200 1200
482632
Multiply the last 2 terms together and then add 1.\nIf that is the pattern (looks like it is) then it should be 155, not 153.\nTo find the next 3 terms, just keep going. (You'll need a calculator with lots of digits).\n22 * 155 + 1 = 3411\n155 * 3411 + 1 = 528706\n3411 * 528706 + 1 = 1803416167
572405
This is the way you find them all\n100 = 1 x 100 = 2 x 50 = 4 x 25 = 5 x 20 = 10 x 10\nFind yourself the dividers of 360 and 300.
568943
17\n1, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 21,31,41,51,61.
627773
1,2,5,10,20,50,100.\n\nSo that's 7.
1103012
35\n\nThe sequence is every even-indexed Fibbonacci number plus 1.  So from the original sequence whose value is the sum of the two previous numbers:\n\n0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, ...\n\nYou take every second number and add 1:\n\n2, 3, 6, 14, 35, 90, 234, ...
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!
318013
What I usually do is;\n•Divide  both numbers into the their  multiples of 10 (in your example (258=200+50) +8) and (63=(60)+3) \n•Then multiply the multiples of the both i.e. (200*60=12000),(50*60=3000) and add (12000+3000=15000) and keep it in your mind. Very easy!\n•Multiply the multiples by residuals i.e. (200*3=600),(50*3=150),\n(60*8=480=500-20(just to make it easier)) and add;\n (600+150 +(500-20)=1250-20=1230) . Add to above (15000+1230=16230) also keep it in mind. Notice that all above numbers are easy to add and easy to remember since they are mostly multiples of 10 (in most examples and not in this one only).\n•Multiply the residuals (Notice that they usually are of 1 digit) here; (8*3=24)\n•Add the last result (usually small quantity) to what you already have in your mind to get the final result. Here in your example (16230+24=16254) which is the answer.\n•If you keep practicing the above method in your spare time by giving any arbitrary examples you would find yourself doing the most elaborated jobs within less a minute. \n\nHope it will work.
1008114
This series is n(n+1)\n\n2,6,12,20,30,42,56,72,90,. . .
627277
(^^you forgot 41)\n25\n\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
209560
P=(V^2)/R\nP=(150^2)/11=2045.45
157712
150 After i answer is question 152
400630
Answer : 64\n\nExplanation : \n2+3=5\n5+5=10\n10+7=17\n17+9=26\n26+11=37\n37+13=50\n50+15=65\n\nSo the next number in the line should be 65, not 64.