Problem - 2806
Let $A_n$ be the average of all the integers between 1 and 101 which are the multiples of $n$ . Which is the largest among $A_2, A_3, A_4, A_5$ and $A_6$?
The smallest multiple of $n$ is of course $n$. Let $a_n$ be the largest multiple of $n$ not exceeding $101$. Then $$A_n = \frac{(n+a_n)}{2}$$
Therefore,
$$
\begin{array}{}
A_2 &=\frac{2+100}{2}&=51\\
A_3 &= \frac{3+99}{2}&=51\\
A_4 &=\frac{4+100}{2}&=52\\
A_5 &=\frac{5+100}{2}&=52.5\\
A_6 &=\frac{6+96}{2}&=51
\end{array}
$$
Hence, the largest one is $\boxed{A_5}$.