Three consecutive integers whose sum is 117 are 38, 39, and 40.
N + (N+1) + (N+2) = 117
3N + 3 = 117
3N = 114
N = 38
117, 118 and 119.
The numbers are 37, 39 and 41.
The square roots of 117 are irrational numbers and so are not two integers - consecutive or otherwise.
38, 39, 40
You can find those by trial and error. You can also write an equation for the three consecutive integers, and solve it. If the first number is "n", the others are "n + 1" and "n + 2". By solving the equation for "n", you get the first of the three numbers.
117/3 = 39, so the three consective odd integers whose sum is 117 are 37, 39, and 41.
(-117)+(-118)+(-119)=(-354)
117, 118 and 119.
117
The numbers are 37, 39 and 41.
The square roots of 117 are irrational numbers and so are not two integers - consecutive or otherwise.
38, 39, 40
These are really simple to solve. Just divide by 3 and that gives you your middle integer. In this case 348/3 = 116 so the trio are 115, 116 and 117.
answer is : 58,59working: x= first number, x + 1= second numberx + x + 1 =1172x= 116x= 58
You can find those by trial and error. You can also write an equation for the three consecutive integers, and solve it. If the first number is "n", the others are "n + 1" and "n + 2". By solving the equation for "n", you get the first of the three numbers.
x is lowest of the integers so x + (x + 2) + (x + 4) = 123 ie 3x + 6 = 123 so 3x = 117 and x = 39, so integers are 39, 41 and 43
39, 41, 43 Let x represent the smallest of these numbers. From the problem, we know x+(x+2)+(x+4)=123 Solving for x: 3x+6=123 3x=117 x=117/3=39 So our integers are: 39, 41, 43