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What is the largest number that can be made using the integers from 1 to 10 under the following conditions:

  • Addition, subtraction, division are allowed.
  • Multiplication and other operations are not allowed
  • All numbers must be used exactly once
  • You can't divide by 0
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Is there anything in the rules preventing us from simply doing

$$ 10 \div (1 \div 2 \div 3 \div 4 \div \cdots \div 9) = 10! = 3628800$$

?

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  • $\begingroup$ Doh! Now why did I not see that! $\endgroup$
    – Jaap Scherphuis
    Commented Sep 4 at 14:14
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    $\begingroup$ @JaapScherphuis IDK, but I observe that not only you but at least two more pretty smart people were kind of fooled by the question. It's fascinating, there seems to be some "implicit structural misdirection" (is that a thing?) embedded in how the problem is posed. $\endgroup$
    – Albert.Lang
    Commented Sep 4 at 14:33
  • $\begingroup$ Ah, of course! Now I'm curious if this is actually optimal (sure seems so, but I don't see an immediate argument) $\endgroup$
    – Tim Seifert
    Commented Sep 4 at 14:56
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    $\begingroup$ I thought I had a proof that this is optimal, but it doesn't quite come together. The idea is to assign a positive score to each rational value such that positive integers equal themselves, and the score for the +, -, or ÷ of two rationals is at most the product of their scores. It seems like f(a/b) = max(a,b) (for non-negative a,b) is such as score, but it turns out that for a/b+c/d = (ad+bc)/ad, the numerator on the right scores too high. $\endgroup$
    – xnor
    Commented Sep 4 at 19:10
  • $\begingroup$ f(a/b) = max(a/b, b/a) also almost works. $\endgroup$
    – Bruno Le Floch
    Commented Sep 4 at 22:47
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I can currently do

35280.

This is achieved via

$\left( 8/\left(\frac{4}{6+1}-\frac{5}{9}\right)\right)/\left(\frac{3}{10}-\frac{2}{7}\right)$, a very complicated way of writing $8\times 63\times 70$.

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