Given that the product, $$15600$$, is a multiple of $$25$$, the three numbers between  them must contribute two factors of $$5$$; since they are consecutive and there are only three of them, this can happen only if one number is itself a multiple of $$25$$. It is now worth observing that the product of three consecutive numbers is roughly the same as the cube of the middle of the three. Since $$20^{3}\textless{} 15 600 \textless{} 30^{3}$$, the middle number must lie between $$20$$ and $$30$$, hence one of the numbers must be $$25$$. The numbers can therefore be $$\left\textbackslash{ {23, 24, 25} \right\textbackslash}$$,~ $$\left\textbackslash{ {24, 25, 26} \right\textbackslash}$$ or $$\left\textbackslash{ {25, 26, 27} \right\textbackslash}$$. The product $$15600 =13 \times1200$$, so it has factors both $$4$$ and $$13$$. The triple $$\left\textbackslash{ {25, 26, 27} \right\textbackslash}$$ has no factor of $$4$$, and $$\left\textbackslash{ {23, 24, 25} \right\textbackslash}$$ no factor of $$13$$. So, by elimination, the numbers are $$24$$, $$25$$ and $$26$$, and their total is $$75$$. 
The product of three consecutive numbers is $$15600$$. What is their sum? 

a) $$75$$ 
b) $$78$$ 
c) $$81$$ 
d) $$84$$ 
e) $$87$$ 