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Sagot :
We are given that the sum of three consecutive numbers is 81. To determine the smallest number we will set "n" as the smallest number, therefore, since the numbers are consecutive this means that the three numbers are:
[tex]\begin{gathered} n=\text{ first number} \\ n+1=\text{ second number} \\ n+2=\text{ third number} \end{gathered}[/tex]Now, since the sum of the numbers is 81 we have:
[tex]n+(n+1)+(n+2)=81[/tex]Now we can add like terms:
[tex]3n+3=81[/tex]Now we will solve for "n", first by subtracting 3 from both sides:
[tex]3n+3-3=81-3[/tex]Solving the operations:
[tex]3n=78[/tex]Now we divide both sides by 3:
[tex]\frac{3n}{3}=\frac{78}{3}[/tex]Solving the operations we get:
[tex]n=26[/tex]Therefore, the smallest number is 26.
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