117
27, 28, 29 and 30.
The numbers are 56 and 58.
The larger is 114, and the other is 112.
114.As we know that the numbers are 2 consecutive even integers, we know that one number will be 2 larger than the other. We can use this to solve the problem with algebra:Where x is the smaller number,x + (x + 2) = 2262x + 2 = 2262x = 224x = 112Therefore the larger integer is 112 + 2 = 114
27 + 28 + 29 + 30 = 114
The consecutive odd integers for 114 are 37, 38 and 39.
118
The numbers are 27, 28, 29 and 30.
117
The integers are 27, 28, 29 and 30.
27+28+29+30=114
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
27 + 28 + 29 + 30 = 114
The integers are 37, 38 and 39.
27, 28, 29 and 30.
27, 28, 29 and 30 is one set that meets the requirement. 37, 38 and 39 is another. There are many such sets.