You have digits from $0$ to $9$, where you need to use every digit only once, and you have only one operator (multiplication, $*$). You may combine/join digits to create a bigger number as shown below:



What is the highest result you can get?

  • $\begingroup$ Is it allowed to use operator * more then once? $\endgroup$
    – z100
    Dec 8, 2016 at 22:48

2 Answers 2


This is not really an independent answer; just an explanation of why the answer given by Levieux is optimal. (Of course it's easy to check by computer, but you might prefer to have a more human-comprehensible explanation.)

First of all, obviously each number's digits must be in descending order because if not, reordering one number's digits to be in descending order increases that number while leaving the other unchanged.

Next, write the two numbers one above the other with most significant digits aligned. (Pad with zeros if you need to.) Then apart from the first digit, the smaller number's digits must be larger than the corresponding digits of the larger number because otherwise swapping two digits for which this isn't true increases the product. (Because if $\delta$ is the difference between the digits taking place value into account, then we are replacing $ab$ with $(a-\delta)(b+\delta)=ab+\delta(a-b-\delta)=ab+\delta(a'-b')$ where $a',b'$ are the new numbers, and we still have $a'>b'$.)

So we have a situation like this:

$\begin{array} \\ 9 & \rightarrow & \bullet & \rightarrow & \bullet & \rightarrow & \bullet & \rightarrow & \bullet \\ \downarrow & & \uparrow & & \uparrow & & \uparrow & & \uparrow \\ \bullet & \rightarrow & \bullet & \rightarrow & \bullet & \rightarrow & \bullet & \rightarrow & \bullet \end{array}$

where arrows mean $\geqslant$ and you should think of the pattern continuing indefinitely to the right where there are infinitely many zeros.

But we could take this diagram and shift the lower number left one place, making it the larger number, and then all but the leftmost vertical arrow would reverse. (Because we would still have a maximal product, so the reasoning above applies "with roles reversed".) In other words, we actually have this (again, continuing infinitely to the right, or at least until whichever number is longer runs out):

$\begin{array} \\ 9 & \rightarrow & \bullet & \rightarrow & \bullet & \rightarrow & \bullet & \rightarrow & \bullet \\ \downarrow & & \uparrow & \searrow & \uparrow & \searrow & \uparrow & \searrow & \uparrow \\ \bullet & \rightarrow & \bullet & \rightarrow & \bullet & \rightarrow & \bullet & \rightarrow & \bullet \end{array}$

and now I want to draw attention to the fact that we have a chain connecting all the digits:

$\begin{array} \\ 9 & & \bullet & & \bullet & & \bullet & & \bullet \\ \downarrow & & \uparrow & \searrow & \uparrow & \searrow & \uparrow & \searrow & \uparrow \\ \bullet & \rightarrow & \bullet & & \bullet & & \bullet & & \bullet \end{array}$

and therefore only one way to fill them in, which is the one in Levieux's answer.

[EDITED because @mathreshka kindly pointed out a small error, which I have now fixed. Thanks!]

  • 1
    $\begingroup$ this is the answer i was looking for. Thanks for detailed explanation. $\endgroup$
    – Oray
    Dec 8, 2016 at 21:43
  • $\begingroup$ Nice, great explanation! I was trying to write down some kind of proof/explanation, but didn't really know where to start. $\endgroup$
    – Levieux
    Dec 8, 2016 at 21:50
  • $\begingroup$ This is excellent, but leaves out one detail; you have not shown that the two numbers should be 5 digits each. $\endgroup$ Dec 8, 2016 at 23:12
  • $\begingroup$ @frodoskywalker: Levieux's answer provides a counterexample to your claim that they have to each be 5 digits. $\endgroup$
    – Ben Voigt
    Dec 8, 2016 at 23:13
  • $\begingroup$ Right. You can move the 0 from one to the other. But what it does show is that the numbers have to be 9642 and 87531 with a zero on the end of one of them. $\endgroup$
    – Gareth McCaughan
    Dec 9, 2016 at 0:15

What about:

$96,420 * 87,531 = 8,439,739,020$

And obviously:

$9,642 * 875,310 = 8,439,739,020$ as well

  • $\begingroup$ Oooh thats higher than ours $\endgroup$ Dec 8, 2016 at 20:20
  • $\begingroup$ You ruined my morning! $\endgroup$
    – YowE3K
    Dec 8, 2016 at 20:21
  • $\begingroup$ This is the highest possible. +1 $\endgroup$
    – Rubio
    Dec 8, 2016 at 20:34
  • 2
    $\begingroup$ I hate to be pedantic, but the actual answer is 9876543210, since OP didn't say the multiplication operator has to be used at all. $\endgroup$ Dec 9, 2016 at 17:12

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