Question

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

Answer 1

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