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So I am helping a grade 5 student to prepare for an olympiad and found this question from the olympiad's past paper :

Using $1, 2, 3, 4, 5, 6, 7$ and $8$ (without repetition), form two three-digit numbers and one two-digit number such that the product of these three numbers is the greatest. Find the sum of these three numbers.

At first glance, it seems really intuitive to directly conclude that it must be 800-something times 700-something times 60-something. But it gets a little bit trickier when it comes to determining the tens place. After trying some values on calculator, I found that the greatest combination is $831 \times 742 \times 65 = 40079130$, but I can't find a suitable and reasonable "rule of thumb" for this question.

So, my point is, are there some thoughts to determining the rule of thumb (not necessarily rigorous) for such question?

Ricky The Ising
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1 Answers1

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One intuition is that the highest digits need to be in the most significant places.

The second is that the three numbers need to be as close together as possible. So once we know which are the four leading digits, this implies that the highest leading digit goes to the shortest number. Then you allocate the second digits so that the longest number with the highest leading digit gets the lowest, and the shortest number with the lowest leading digit gets the highest.

So we go leading digits $8,7,6$ and allocate them $8*, 7**, 6**$

Then second digits $5,4,3$ and allocate them $83, 74*, 65*$

Then allocate $2,1$ to get $83, 741, 652$

Note - as in one of the comments below, if you try to make the numbers close before you make them big, you might get an outcome like $712$ for one of the numbers, but then $721$ is larger and will give a higher product. $1$ always needs to be allocated to the least significant digit.

Mark Bennet
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    very good second observation that numbers should be as close as possible to maximize the product – Vasili Jul 26 '18 at 17:03
  • "Then you allocate the second digits so that the longest number with the highest leading digit gets the lowest" Isn't that $71\cdot$ ? I know you answer is right, but is the explanation? – Ander Biguri Jul 26 '18 at 19:48
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    @AnderBiguri The first thing comes first. So with four numbers to multiply you would take the four highest numbers for the most significant digits. Then the next four highest would be the second digits (unless some one digit numbers were involved) - that takes priority. Once that allocation is done, the digits chosen are allocated to make the numbers as close as possible. – Mark Bennet Jul 26 '18 at 19:52
  • Certainly is right, just letting you know that maybe the explanation is not 100% obvious at first read. Great answer anyway. – Ander Biguri Jul 26 '18 at 19:55
  • @AnderBiguri I've made some changes to try to help. – Mark Bennet Jul 26 '18 at 20:08