<s>[INST] Divisors of $2^kp$ Let $p$ be an odd prime number. What is the necessary and sufficient condition (in terms of $p$ and $k$) such that we can partition the divisors of $2^kp$ into two set with equal sum. 
 [/INST] Let $N=2^kp$. The divisors of $N$ are $1,2,4,\dots, 2^k$ and $p,2p,4p,\dots, 2^kp$. Their sum simplifies to $(2^{k+1}-1)(p+1)$.
If we divide the divisors of $N$ into two sets, then $N$ will be in one of the sets. The sum of the rest of the divisors is $(2^{k+1}-1)(p+1)-2^kp$. This simplifies to 
$$N +(2^{k+1}-1-p).$$
If the term $2^{k+1}-1-p$ is negative, then the divisors less than $N$ add up to less than $N$, so we cannot do the required splitting. 
If $2^{k+1}-1-p$ is $\ge 0$, then we can do the splitting. In the case $2^{k+1}-1-p=0$, the number $N$ is a  perfect number. The splitting consists of $N$ as the only member of one of the sets, and all the  rest of the divisor of $N$ in the other set. 
Now we verify that we can do the splitting also in the case $2^{k+1}-1-p\gt 0$. 
If we put just $N$ into set $A$, and the rest of the divisors into set  $B$, then set $A$ has sum $N$ and set $B$ has sum $N +(2^{k+1}-1-p)$. So we must "transfer" from $B$ to $A$ a collection of numbers with sum $\frac{2^{k+1}-1-p}{2}$.
This is easy to do. For the number $\frac{2^{k+1}-1-p}{2}$ is less than $2^{k+1}-1$. And any number $\le 2^{k+1}-1$ can be expressed as a sum of distinct powers of $2$ each $\le 2^k$. These powers of $2$ are among the divisors of $N$, so by shifting suitable divisors of $N$ from $B$ to $A$ we get the desired splitting. 
There is even a mechanical method of doing it: just write down the binary representation of $\frac{2^{k+1}-1-p}{2}$.
Conclusion:  The divisors of $2^kp$ can be split into two sets with equal sum if and only if $2^{k+1}-1-p\ge 0$, or equivalently if and only if $p\le 2^{k+1}-1$. 
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