Maximize the integral product

Given $a_i \in \mathbb{Z}^+ \forall 1\le i \le n$ , and

$a_1 + a_2 + \cdots + a_n = m$ , where $m$ is given positive integer and

$a_1 a_2 \cdots a_n$ is maximum.

Find $a_i, n$ for a given integer $m$ .

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Note that $2(k-2) > k$ for $k > 4$ , that $k+1 > k\times1$ for $k \ge 1$ , and that $3^2 > 2^3$ .

Firstly, note that there are finitely many ways of writing $m$ as a sum of positive integer (there are at most $m$ integers in the sum, and each is no greater than $m$ ). Thus there certainly is a maximum possible product over all such sums (since there are only finitely many products to consider).

If one of the numbers $a$ in a decomposition of $m$ is greater than $4$ , then we can replace $a$ by $2$ and $a-2$ , obtaining a new decomposition of $m$ with a larger product. Thus there are no integers greater than $4$ in the decomposition of $m$ that yields the maximum product.
If one of the numbers $a$ in a decomposition of $m$ is equal to $1$ , another number in the decomposition is equal to $b$ , replacing the pair of numbers $a=1,b$ by the single number $b+1$ yields a new decomposition of $m$ with a larger product. Thus there are no integers equal to $1$ in the decomposition of $m$ that yields the maximum product.
Since $4 = 2+2=2\times2$ , any copy of the integer $4$ can be replaced with two copies of $2$ , so there is no need for any decomposition of $m$ to contain the number $4$ .

Thus the decomposition of $m$ that yields the maximum product contains only the numbers $2$ and $3$ . Since $2^3 < 3^2$ and $2+2+2=3+3$ , any set of three $2$ s can be replaced by two $3$ s, resulting in a larger product. Thus the decomposition of $m$ that yields the maximum product cannot have more than two $2$ s.

If $m = 3k$ , the only way we can satisfy all the above requirements for a decomposition (only $2$ s and $3$ s, and at most two $2$ s) is to have $k$ copies of $3$ , making the maximum product $3^k$ .
If $m = 3k+1$ , the only way to satisfy these rules is to have two $2$ s and $k-1$ copies of $3$ , making the maximum product $2^2 \times 3^{k-1}$ .
If $m = 3k+2$ we are forced to have just one $2$ and $k$ copies of $3$ , making the maximum product $2\times3^k$ .

Mark Hennings - 4 years ago

@Pi Han Goh @Mark Hennings

Kartik Sharma - 4 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$