Can we identify the largest product of two numbers made from four given digits My problem is, given 4 positive integers (single digits 0-9), is there a generic way to identify which combination of them will yield the largest product of a single multiplication?
So to clarify if my digits were: 3, 8, 1, 6
I think the largest number I could make given a single multiplication would be 8 * 631 = 5048
or in generic terms, the largest single digit multiplied by a combination of the rest of the digits in a sequence of decreasing magnitude. 


*

*is that correct?

*is there a way to formally state that? Sort of a proof.

 A: It should be obvious that the digits in each number must be decreasing.  It looks like you prohibit $abcd (\times 1)$.  We can sort the digits $a \ge b \ge c \ge d$ - do you allow pairs to match?  First let's look at splitting the digits $3$ and $1$.  We need to compare $a \times bcd, b \times acd, c \times abd, d \times abc$.    Comparing $a \times bcd=100ab+10ac+ad$ with $b \times acd=100ab+10bc+bd$ we see the former is larger because $a \ge b$  A similar argument will show that $a \times bcd$ beats the others.
Similarly we can compare $ac\times bd=100ab+10(ad+bc)+bd$ with $ad \times bc=100ab+10(ac+bd)+cd$.  We would rather have more copies of the big $a$, so the second wins.
Finally we need to compare $a \times bcd=100ab+10ac+ad$ with $ad \times bc=100ab+10(ac+bd)+cd$.  The second is the winner.  
A: We have digits
$a, b, c, d, e, f, g, h, i \in S=\left\{ 1, 2, 3, 4, 5, 6, 7, 8, 9 \right\}$,
$a \geq b \geq c \geq d \geq e \geq f \geq g \geq h \geq i$, 
and we want to find out the maximum possible product of two numbers, constructed by digits from the set $S$, by each time having one more extra digit starting from two digits to use option, then we would have the following
for the case $a>b$ we have:
$a \cdot b=max$,
$a \cdot \overline{bc}=max$,
$\overline{ad} \cdot \overline{bc}=max$,
$\overline{ad} \cdot \overline{bce}=max$,
$\overline{adf} \cdot \overline{bce}=max$,
$\overline{adf} \cdot \overline{bceg}=max$,
$\overline{adfh} \cdot \overline{bceg}=max$,
$\overline{adfh} \cdot \overline{bcegi}=max$.
With line over the number meaning $\overline{bce}=b\cdot 10^2+c\cdot 10+e$, representing three digit number with digits $b, c$ and $e$ respectively, which is different from $bce=b\cdot c\cdot e$ etc.
There is a clear pattern there. Each time adding a new digit in a zik-zak order to one of the factors, starting from "$b$" in three digit option, as its new last digit.
In case $a=b$ we ignore $a$ and $b$ for a moment, do the same procedure as above with rest of the optional digits and then just add $a=b$ as first digit to each of the factors.
This can be done with any number of digits from the set $S$. 
There will be some equal digits when you have more then nine of them to use but the zik-zak method from above is the same to get the maximum product.
To make sure those are the maximum products you could check $2^{n-1}-1$ possible potential maximum products for $n$ optional digits from the set $S$. I did that for the case of $n=5, 6, 7$, and wanted to do it for the $n=8, 9$ also, but there is so much work to do there and you get lost easily. You could probably use Mathematica or some other program to do the calculation but I haven't done that. So I searched for another shorter approach and seem to find it. 
This can be better understood if you put both factors one over the other and then multiply them in a "criss-cross" method known from vedic math, also known as "vertically and cross-wise" method, which is not thought in school usually. You could find it in this Book, pages $41$ to $44$.
There is this same problem mentioned in this Book, for the nine digit option case, pages $5$ and $63$. There is also few direction points for explanation there and the case with construction of two factors one over the other which could be connected with the "criss-cross" method.
A: The following algorithm will yield two numbers with the maximum product for any number of given digits, possibly with repetitions.

*

*Start with two "numbers" $x, y$ with no digits.

*Go through all given digits in descending order one at a time. For each digit add it to the smaller of numbers $x, y$. A number with no digits has value $0$. If $x=y$ add the digit to any one of them.

Examples:
$4, 3, 2, 1\;\;\;ad \times bc\;\;\;41 \times 32 = 1312$
$5, 4, 3, 2, 1\;\;\;ad \times bce\;\;\;52 \times 431 = 22412$
$4, 4, 3, 2, 1\;\;\;ac \times bde\;\;\;43 \times 421 = 18103$
Notice different alternation patterns between the last two examples.
