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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.

  1. is that correct?
  2. is there a way to formally state that? Sort of a proof.
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$83\cdot61 = 5063$. –  MJD Jan 22 '13 at 14:35
    
$81*63 = 5103 > 5048$. I suspect this gives the highest product: the larger two numbers as tens digits, then the smaller two numbers as ones digits, but switched (so the smallest digit goes with the largest digit). –  Jonathan Christensen Jan 22 '13 at 14:35
    
@MJD - well, I guess that answers #1. –  Mike Jan 22 '13 at 14:36
    
And $8631$ is larger yet, taking the empty multiplier to be $1$ as is often done. –  Ross Millikan Jan 22 '13 at 14:43
    
@RossMillikan - What is an empty multiplier? Keeping in mind I'm no math major... –  Mike Jan 22 '13 at 14:46

1 Answer 1

up vote 3 down vote accepted

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.

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Perfect! I assume we could do this for any number of digits the proof would just get longer as there are more combinations to check? –  Mike Jan 22 '13 at 15:01
    
@Mike:Yes. I believe the answer is that you split the number of digits as evenly as possible and put all the big ones behind the second biggest. So if you have $a,b,c,d,e,f$ in that decreasing order you want $aef \times bcd$ –  Ross Millikan Jan 22 '13 at 15:07
    
@RossMillikan It seems to me you want to alternate, so you would have $ade \times bcf$ in the case of siz digits. For example, if you have 1-6, yours gives $621\times543 = 337203$, while mine gives $632\times541 =341912$. –  Jonathan Christensen Jan 22 '13 at 17:32
1  
@JonathanChristensen: you are right, you want the third and fourth largest in the tens place. But you want the larger of each pair behind the smaller leading digit. Here $631 \times 542=342002$ This lets you multiply the $2$ by a larger number. –  Ross Millikan Jan 22 '13 at 17:48

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