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Sum of some positive integers is 3841. If P is the product of those numbers, what is the maximum value of P?

Briefly explain.

1
by TPS

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Briefly explain.

by TPS

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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.

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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.

So the

The

You can check up the product of any other partition. It should be less than this.

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I hope the following

Let us divide the number (an integer sum) S as :

=>

Differentiate partially wrt x1, then partially wrt x2, then partially wrt x3 and so on.. Equate them to 0 for maximization.

=>

=>

Similarly when we do dP/d x2, we get

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.

To find the optimum value of n, differentiate p wrt to n:

=> 1/P * d P/ dn = Ln S - n * 1/n - Ln n

=

Equating the derivative to 0, we get

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.