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

User Sephora
by
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1 Answer

1 vote

Answer:

38, 39, and 40

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.

User Jaquan
by
5.7k points