The square roots of 117 are irrational numbers and so are not two integers - consecutive or otherwise.
The numbers are 38, 39, and 40.
The numbers are 37, 39 and 41.
The first ten consecutive composite positive integers are: 114 115 116 117 118 119 120 121 122 123
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
If you mean two consecutive odd numbers then they are 117 and 119
The square roots of 117 are irrational numbers and so are not two integers - consecutive or otherwise.
The numbers are 38, 39, and 40.
The numbers are 37, 39 and 41.
117/3 = 39, so the three consective odd integers whose sum is 117 are 37, 39, and 41.
The first ten consecutive composite positive integers are: 114 115 116 117 118 119 120 121 122 123
117
Let the smallest number be x. Then the others are x+1, x+2 and x+3 and x + (x+1) + (x+2) + (x+3) = 474 iff 4x = 468 iff x = 117, So the consecutive whole numbers 117, 118, 119 and 120 sum to 474.
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
9 and 6 9-6 = 3 92+62 = 117
117, 118 and 119.
Well the factors of 117 are: 1, 3, 9, 13, 39, and 117 itself. Therefore: 1x117 = 117 3x39 = 117 9x13 = 117