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