The sum of four consecutative odd integers is 152.Find the four integers.?
79622
let "x" = first odd integer, therefore second number must be "x+2", third must be "x+4", and so on with the fourth.  set these equal to 152, solve for "x", then solve for remaining numbers:\n\nx+(x+2)+(x+4)+(x+6) = 152\n\n4x+12 = 152\n\n4x = 140\n\nx = 35\n\ntherefore the four numbers are 35, 37, 39, and 41.\n\n35+37+39+41 = 152 :)
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
200692
15,21,25,28,29,5,1,3,99,31
88384
1 * 1650\n2 * 825\n3 * 550\n5 * 330\n6 * 275\n10 * 165\n11 * 150\n15 * 110
361589
Sum of all numbers 1 to 100:\nS = n[2a1 + (n - 1)d]/2\n= 100[2(1) + (100 - 1)]/2\n= 100(101)/2\n= 5050\n\nSum of all numbers 3 to 99, multiples of 3:\nS = n[2a1 + (n - 1)d]/2\n= 33[2(3) + (33 - 1)3]/2\n= 33(6 + 96)/2\n= 1683\n\nSum of all numbers 5 to 100, multiples of 5:\nS = n[2a1 + (n - 1)d]/2\n= 20[2(5) + (20 - 1)5]/2\n= 20(10 + 95)/2\n= 1050\n\nSum of all numbers 15 to 90, multiples of 15:\nS = n[2a1 + (n - 1)d]/2\n= 6[2(15) + (6 - 1)15]/2\n= 6(30 + 75)/2\n= 315\n\nSum of integers from 1 to 100 which are not divisible by 3 and 5:\nS = sum(1-100) - sum(3-99) - sum(5-100) + sum(15-90)\n= 5050 - 1683 - 1050 + 315\n= 2632\n\nWe have to add in the sum(15-90) because in removing sum(3-99) and sum(5-100), we're taking out multiples of 15 twice.  Adding sum(15-90) ensures that we only take out 15 once, 30 once, etc.
630532
guess if algebra, it would go like this 2 minus make a plus \nso 14- - (+) 14=28/14= 2
827033
36=36 (6^2)\n100,000=100,000 (10^4)\n196=196 ( 14^2)\n512=512 (2^9)\n81=81 (9^2)
306814
204 total.\nThe size breakdown is as follows:\n204 =1+4+9+16+25+36+49+64 = n(n+1)(2n+1)/6 when n=8 \n\n1 size 8x8\n4 size 7x7\n9 size 6x6\n16 size 5x5\n25 size 4x4\n36 size 3x3\n49 size 2x2\n64 size 1x1
1025267
Here is how:\n\n6+6+3\n6+5+4\n5+5+5\n\nnow change the first two around until you have all the possiblities.
1092764
4+4 will = 1\n\nThe logic is the answers are going down from five only in odd numbers thus the answer is 1.
365053
It's 15\n\n3 + 3 + 3 x 3 \n(3 + 3) + (3 x 3)\n(6) + (9) = 15
300260
Alright, I've got two answers using the four basic operations:\n\n(14 - 6) * 3\n\nand\n\n(6/(1-(3/4)))\n\nIf you allow other operations, there are a few more answers (! is the factorial symbol : 4! = 4*3*2*1, 3! = 3*2*1, etc)\n\n4*1*sqrt(36)\n(3*6-14)!\n16*3-4!
908655
Let the integers be 'x' and 'x + 2'\nx + x + 2 = 114\nx + x = 114 - 2\nx + x = 112\n2x = 112\nx = 112/2\nx = 56\nThe two integers are 56 and 58.\n\nHope that helped.
403642
just remeber, if you want to know if any number is divisible by 4 just check to see if the last two digets are divisible by 4... that should greatly help in solving this problem\n\nNow if you would please, stop asking all of your homework questions
135926
15 + 30 = 45\n45 x 3 = 135\n45 + 135 = 180
904203
Let the two integers be 'x' and 'x+2'\nx + x + 2 = 4^3\n2x + 2 = 64\n2x = 64 - 2\n2x = 62\nx = 62/2\nx = 31\nThe integers are 31 and 33.
424362
Problem #1:  Four consecutive integers have sum of -34.  Let the first integer be x.  The next three must be x+1, x+2, and x+3.  This gives you the equation\n\n-34 = x + x+1 + x+2 + x+3\n-34 = 4x + 6\n-40 = 4x\nx = -10\n\nSo, the 4 consecutive integers are -10, -9, -8, and -7.\n\nProblem #2:  Again, four consecutive integers can be written as x, x+1, x+2, and x+3.  This time the equation is:\n\n4(x + x+3) = 24 + 6(x+2)\n4(2x + 3) = 24 + 6x + 12\n8x + 12 = 6x + 36\n2x = 24\nx = 12\n\nSo, the 4 integers are 12, 13, 14, and 15.  \n\nIt pays to check your answers:  The sum of the first and fourth is 12+15=27.  Four times this sum is 108.  6 times the third integer is 6*14=84.  108 is 24 greater than 84.