Question

Pearl writes down seven consecutive integers, and adds them up. The sum of the integers is equal to 21*4 times the largest of the seven integers. What is the smallest integer that Pearl wrote down?

Answer 1

Let the smallest integer be $n$. Then the seven consecutive integers she wrote down are $n$, $n+1$, $n+2$, $n+3$, $n+4$, $n+5$, and $n+6$. Their sum is $$n+(n+1)+(n+2)+(n+3)+(n+4)+(n+5)+(n+6)=7n+21.$$The largest of these integers is $n+6$, so we can write the given equation as $$7n+21=21\cdot 4(n+6).$$Simplify this equation to get $7n+21=84n+504$, which means $77n=483$ and $n=\boxed{6}$.