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Find three consecutive integers whose sum is 216. (Hint: if n represents the smallest of the three integers, n + 1 and n + 2 represent

the other two numbers.)

User HAS
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1 Answer

3 votes

Answer:

71, 72, 73

Explanation:

So we want to find three consecutive integers that equal 216.

Let the first integer be n.

Then the second integer is n+1, and the third integer is n+2.

Thus:


n+(n+1)+(n+2)=216

Combine like terms:


3n+3=216

Subtract 3 from both sides:


3n=213\\

Divide 3 from both sides:


n=71

So, the first term is 71.

And the other two is 72 and 73.

And we are done :)

User Matthews
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