The numbers are 31, 32 and 33.
-30 -31 -32
32+33+34=99
-30,-31,-32
Average is 96/3 ie 32 so integers are 30, 32 and 34.
31 32 33
The numbers are 31, 32 and 33.
31, 32 and 33
-30 -31 -32
-31, -32, -33
32+33+34=99
-30,-31,-32
That isn't possible; three consecutive integers, or three consecutive positive integers, always have a sum that is a multiple of 3. In general, you can solve this quickly by trial and error. In this case, you will quickly find that a certain set of three consecutive integers will give you a sum that is TOO LOW, while the next-higher even integers will give you a sum that is TOO HIGH. You can also write an equation and solve it: n + (n + 2) + (n + 4) = 32. If you solve it, you will find that the solution is fractional, not integral.
Average is 96/3 ie 32 so integers are 30, 32 and 34.
The integers are 15 and 17.
If three consecutive integers have the sum of 96, then the problem can be expressed with the equation... N + (N+1) + (N+2) = 96 ...Simplify that and solve and you get... 3N + 3 = 96 3N = 93 N = 31 ... so the three integers are 31, 32, and 33.
30 + 31 + 32 = 93 ( I forgot how to find this algebraically )