combinatorics - Partitioning an integer into $k$ equal parts

Say I have an integer $M$. Is there a one-line command to create a partition of $M$ into $k$ integers s.t. the difference between any two integers is as small as possible?

For example, with $M = 100$ and $k = 10$, we would create the partition: {10,10,10,10,10,10,10,10,10,10}. However, for $k = 7$, we might have the partition: {14,14,14,14,14,14,16}, or better {14,14,14,14,14,15,15}.

For a partial solution, you can of course write:

M = 100;
k = 7;
BalancedPartition = Array[Floor[M/k] &, k];
BalancedPartition[[k]] += M - k*Floor[M/k];

BalancedPartition

{14, 14, 14, 14, 14, 14, 16}

Answer

I do not know single command to do this (it does not imply it does not exist:)). But it is two-line command:

dec[val_, par_] :=With[{ip = IntegerPart[val/par], md = Mod[val, par]}, 
                       ConstantArray[ip, par-md]~Join~ConstantArray[ip+1, md]]


dec[100, 10]

{10, 10, 10, 10, 10, 10, 10, 10, 10, 10}

dec[100, 7]

{14, 14, 14, 14, 14, 15, 15}

This is straightforward method: for k-integer components md=Mod[val,k] ε [0, k-1] so it is adding 1 to md last positions of final list.

One could use Quotient[val,par] or Floor[val/par] insted of IntegerPart.

Edit. Also:

ConstantArray in place of Table will make this faster. – Michael E2

So I swapped them.