4
$\begingroup$

Given $A = B \times C$

Using any $9$ digits form two numbers $B$ and $C$.All digits must be used exactly once.

What is the maximum possible value of $A$?

  • 0
    Morgan: So you want to maximize $C = A/B$ where $A$ and $B$ are $m$ and $n$ digits respectively?2011-01-22

3 Answers 3

0

Instead of "trying all possibilities", one can try the following approach (which I haven't). Consider some current assignment for $B,C$, and try to modify it somehow - change the location of a single digit. Find some condition ensuring that $(B,C)$ is a local maximum. Enumerate all local maxima, and take the maximal one (if we're lucky, there's only one local maximum, which must be global).

Using this approach, we can verify for example that the digits in each number must come in decreasing order (from MSD to LSD).

16

If you have a cord and you want to encircle the biggest area possible with it, I'm sure you're well-aware of the fact that making a circle is the way to go to optimize that space. However, if we are under the same constraints as here, it has to be a rectangle, in which case the most optimal solution would be a square.

It doesn't take long to convince yourself that $n*n$ always will be bigger than $(n-1)*(n+1)$.

Therefore, the most optimal solution would be having the two numbers be as close to each other as possible.

This is done by repeatedly taking the biggest digit and assigning it as a suffix to the smallest of your 2 new numbers $B$ and $C$.

E.g.: $\{0, 0\} \rightarrow \{9, 8\} \rightarrow \{9, 87\} \rightarrow \{96, 87\} \rightarrow \{96, 875\} \rightarrow \space ... \space\rightarrow \{9642, 87531\} $

  • 0
    However, it'd appear that you are right that the algorithm wouldn't work, but the issue is in how the numbers are distributed. [I'll refer future readers to this answer](https://math.stackexchange.com/a/2900799/11781).2018-09-04
3

By brute force (Mathematica): 843,973,902 = 87,531 • 9,642.

Probably not the most elegant way to do it, but here's my code:

In[21]:= Sort[  Table[Sort[     Module[{a = FromDigits[#[[1 ;; 9 - d]]],          b = FromDigits[#[[10 - d ;; 9]]]}, {a b, {a, b}}] & /@       Permutations[{1, 2, 3, 4, 5, 6, 7, 8,         9}], #1[[1]] > #2[[1]] &][[1]], {d, 1, 4}], #1[[1]] > #2[[1]] &]  Out[21]= {{843973902, {87531, 9642}}, {843809444, {875321,     964}}, {840414816, {8754321, 96}}, {788888889, {87654321, 9}}} 
  • 0
    @PEV: Issac understood the question right, I myself got it wrong before.2011-01-22