Question

The sum of 3 consecutive integral numbers is 117. Find the numbers.

Answer 1

Answer:

38, 39, and 40

Step-by-step explanation:

Let the three (3) consecutive integral numbers be:  

n  

n + 1  

n + 2  

Add the numbers together:  

n + n + 1 + n + 2  

Combine like terms:  

n + n + n = 3n  

1 + 2 = 3  

Therefore:  

3n + 3 = 117  

Subtract 3 from each side of the equation.  

3n + 3 - 3 = 117 - 3  

 

Combine like terms:  

3 - 3 = 0  

117 - 3 = 114  

3n = 114  

Divide each side by 3  

3n/3 = 114/3  

n = 38  

n + 1 = 38 + 1 = 39  

n + 2 = 38 + 2 = 40  

Check: 38 + 39 + 40 = 117  

The three (3) consecutive integral numbers are:  

38, 39, and 40.