The some of three consecutive integers is 33 more than the least of the integers. Find the integers

Let the smallest number = x

The next higher number would be x + 1

The 3rd number would be x + 2

The sum of the 3 numbers: x + x+1 + x+2

33 more than the smallest number is written as x+33

Now you have:

x + x+1 + x+2 = x+33

Simplify the left side:

3x + 3 = x + 33

Subtract 3 from each side:

3x = x + 30

Subtract 1x from each side:

2x = 30

Divide both sides by 2:

X = 30/2

x = 15

The three numbers are 15, 16 and 17