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.
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.
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
117/3 = 39, so the three consective odd integers whose sum is 117 are 37, 39, and 41.
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.
(-117)+(-118)+(-119)=(-354)
The square roots of 117 are irrational numbers and so are not two integers - consecutive or otherwise.
10 and 11 or -11 and -10
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
38, 39, 40
117 = 3*39, a product of two integers which are larger than 1.
117 = 3*39, a product of two integers greater than 1.
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