Not A Perfect Number

$A = 2^n\cdot (2^{n+1}-1)$ is a perfect number if and only if $2^{n+1}-1$ is prime. Thus,

$\begin{array}{c|rl|rl} n & 2^{n+1}-1 & & 2^n\cdot (2^{n+1}-1) \\ \hline 1 & 3 & \text{is prime} & 6 & \text{is perfect} \\ 2 & 7 & \text{is prime} & 28 & \text{is perfect} \\ 3 & 15 & \text{isn't prime} & 120 & \text{is imperfect} \\ 4 & 31 & \text{is prime} & 496 & \text{is perfect} \\ 5 & 63 & \text{isn't prime} & 2016 & \text{is imperfect} \\ \vdots & & \vdots & & \vdots \end{array}$

Thus the answer is $\boxed{2016}$ . It may not be a perfect year, but it is special for sure!

We can use Euler's equation for $perfect$ $numbers$ .

$\boxed { 2 ^{p-1} (2^{p} -1) }$

where $p$ and $2^{p}-1$ both are $prime$ $numbers$ .

Applying $p-1$ = $n$ we get a perfect number for $n$ = 4 Therefore we find that for $n$ =5 , $p-1$ is not a prime number. So we can present smallest imperfect number larger than 120 as following

$\boxed { 2 ^{5} (2^{6}- 1) }$

= $\boxed {2016}$ .

One of the most beautiful problems I have seen on brilliant. didn't come to my mind how to solve it using theory. but it became very very easy using programming.