Five consecutive integers

Statement

The sum of five consecutive integers is $22$ more than the largest of the five. What is the smallest one?

Solution

Unknown. Let $n$ be the smallest of the five integers. The others are $n+1,\,n+2,\,n+3,\,n+4$, and the largest is $n+4$.

Equation. The sum of the five integers is $n+(n+1)+(n+2)+(n+3)+(n+4) = 5n+10.$ This is $22$ more than the largest: $5n+10 = (n+4)+22 \implies 5n+10 = n+26 \implies 4n = 16 \implies n = 4.$ Check: the integers are $4,5,6,7,8$, their sum is $30$, and $30-8=22$. Correct. $\boxed{n = 4}$