Divisors of $2^kp^r$ Let $p$ be an odd prime number. What is the necessary and sufficient condition (in terms of $p$ and $k,r$) such that we can partition the divisors of $2^kp^r$ into two set with equal sum. You may want to LOOK HERE where the special case $2^kp$ is dealed. 
 A: For $r$ even, the sum of divisors is odd, hence such a partition is never possible.
For odd $r$ with $2^{k+1}\ge p+1$ consider the following procedure:
Take all divisors $d$ of $N=2^kp^r$, sort them from highest to lowest as $d_1=N>d_2>\ldots >d_m=1$. Then $d_{i+1}\ge \frac12d_i$ because for $d_i=2^ap^b\ne 1$ we either have $a\ge 1$ and hence $d_{i+1}\ge 2^{a-1}p=\frac12d_i$ or $d_i=p^b$ with $b\ge 1$ and then one of the numbers $2p^{b-1},\ldots, 2^kp^{b-1}$ is in $[\frac12 d_i,d_i)$ because $2^{k+1}p^{b-1}>d_i>2p^{b-1}$.
Now place the divisors into "pots" $A$ and $B$ from biggest to smallest divisor, each time selecting the pot with smaller sum. By induction, the differences between the pots is at most as big as the last number added:
After the first step (putting $d_1=N$ into one of the pots), the difference is exactly $d_1=N$. If after adding $d_i$ the difference is $\delta\le d_i$, then the new difference after adding $d_{i+1}$  will be $|\delta-d_{i+1}|\le \max\{d_i-d_{i+1},d_{i+1}\}=d_{i+1}$ because $d_i\le 2d_{i+1}$.
Therefore, the final difference is at most $d_m=1$. Since the difference must be even, we conclude that the difference is zero.
For odd $r$ with $2^{k+1}< p+1$ assume there is a partition as desired.  Drop all summands not divisible by $p$ from both partitions; this creates a difference of at most $2^{k+1}-1<p$ between the partitions. Now divide all remaining summands by $p$; this divides the difference by $p$ as well, which must therefore be $<1$, hence zero. But now we have a partition of all divisors of $2^kp^{r-1}$ into equal sums, which is impossible (see above, even case). 
Cunclusion. If $p$ is an odd prime, then the divisors of $2^kp^r$ can be partitioned into two sets of equal sum if and only if $r$ is odd and $2^{k+1}>p$.
