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
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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.