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Question 623586: The sum of 3 consecutive integral numbers is 117. Find the numbers.
Answer by lenny460(1073)   (Show Source):
You can put this solution on YOUR website! 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.
obuong3@aol.com
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