The sum of three numbers a, b, and c is 99. If we increase a by 6, decrease b by 6 and multiply c by 5, the three resulting numbers are equal. What is the value of b?
$a+6=b-6=5c\implies a=b-12, c=\frac{b-6}{5}$
$a+b+c=99\implies b-12+b+\frac{b-6}{5}=99\implies \frac{11}{5}b-\frac{66}{5}=99\implies \frac{1}{5}b-\frac{6}{5}=9\implies b-6=45\implies b=\boxed{51}$.