Question

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.)

Answer 1

Answer:

71, 72, 73

Step-by-step 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 :)