117
-1, -3, -5
Start with "-3", then add one at a time to get as many consecutive integers as you want.
There are no consecutive odd integers whose sum is equal to 35. 9+11+13 = 33, and 11+13+15 = 39.
It need not be. 2 and 3 are consecutive integers and their sum is 5, not 31.
114/3 = 38, so the three integers are 37, 38 and 39.
117
-1, -3, -5
27, 28, 29
36, 37 and 38.
They are 1986, 1987 and 1988.
129 divided by 3 is 43, so the 3 consecutive integers having a sum of 129 are 42, 43, and 44.
They are: 57+58+59 = 174
65/67/69
The numbers are 121, 122 and 123.
114, 115 & 116
Since the average of the three integers will be 114/3 = 38, and the three numbers are consecutive, the numbers will be 36, 38 and 40. Another way to do this problem using algebra is to let the first integer be n, then the next two are n+2 and n+4. Their sum is 3n+6 and it equals 114 So 3n+6=114 and 3n=108 so n=36 then next two numbers must be 38 and 40 since they are consecutive even integers.