This is my own creation/idea. I Hope this is valid universally for all numbers like 3841 above.
Divide the number in to N integers such that each part is nearly equal to e (Euler number 2.718..). The product will be the maximum. Since the closest integer number to e is 3. Divide 3841 into 1279 parts of value 3 each and two more parts of value 2. After 3, the nearest integer to e is 2.
So the sum is: 3 + 3 + 3 + .... 1279 times + 2 + 2 = 3831
The product of these numbers which is maximum for any partition of 3841:
3¹²⁷⁹ * 2 * 2 ≈ 6.9 * 10⁶¹⁰
You can check up the product of any other partition. It should be less than this.
I hope the following generic solution for any integer sum S is formal and is a valid proof. I give the proof here below.
Let us divide the number (an integer sum) S as :
x1 + x2 + x3 + x4 + .... + xn = S where 1<= x_i <= S
=> xn = (S - x1 - x2 - x3 - .... x_n-1)
Product: P = x1 x2 x3 x4 .... x_n-1 * (S - x1 - x2 - x3 .... - x_n-1) = S
Differentiate partially wrt x1, then partially wrt x2, then partially wrt x3 and so on.. Equate them to 0 for maximization.
dP/d x1 = (x2 x3 x4 ...x_n-1) [S - 2 * x1 - x2 - x3 ....- x_n-1] = 0
=> 2 x1 + x2 + x3 + .... + x_n-1 = S
=> x1 = xn
Similarly when we do dP/d x2, we get x2 = xn and so on...
So the product is maximum when all parts (partitioned parts) are equal.
Now we have to find out the number "S/n" of each part and number "n" of parts in to which S is to be divided.
Sum = S = n * S/n
Product = P = (S/n)^n = S^n / n^n
To find the optimum value of n, differentiate p wrt to n:
Ln P = n Ln S - n Ln n
=> 1/P * d P/ dn = Ln S - n * 1/n - Ln n
= Ln (S/n) - 1
Equating the derivative to 0, we get S/n = e (Euler e = 2.718..)
Thus each part is equal to e and there are S/e parts into which S is divided.
Since this question is integer based, we will have parts equal to 3 if the number is completely divisible by 3. Otherwise, we will have either one part or two parts of value 2.