Three consecutive integers have a sum of 96. Find the smallest of them.

Step-by-step solution

1. 1

   Let the variable stand for the smallest integer. Then the three are $x$, $x+1$ and $x+2$. Choosing the smallest as $x$ means the final answer will come straight out of the equation with no extra step.

2. 2

   Set up the equation.

   $x+(x+1)+(x+2)=96$

3. 3

   Solve.

   $3x+3=96\;\Longrightarrow\;3x=93\;\Longrightarrow\;x=31$

4. 4

   Read off the requested value. The integers are $31$, $32$, $33$, and the smallest is

   $\boxed{31}$

5. 5

   Check, and cross-check with the mean. $31+32+33=96$ ✓. The average is $96\div 3=32$, which must be the middle integer for evenly spaced numbers, so the smallest is $32-1=31$ — confirming the algebra. Compare the sister problem with sum $72$, where the largest was wanted and the answer was $x+2$ rather than $x$; always re-read which of the three the question asks for.