Logic Puzzle: Pandigital e

If you can construct a number using the digits, 0,1,2,3,4,5,6,7,8, and 9 each exactly once and the operations of addition, subtraction, multiplication, division, exponentiation, and digit concatenation, in any order, we will call that number pandigital. Notice that there are only finitely many pandigital numbers.

How close can you get to the mathematical constant e =\sum_{n=0}^\infty \frac{1}{n!} with a pandigital number?

There are two parts to this logic puzzle. First, try to get the best approximation of e that you can. Then, try to come up with an estimate of how close you can get if you checked all possible pandigital approximations. I will post one particularly good approximation in the comments, but I suggest you think about it for a while before looking.



Read 6 comments

  1. You can actually get at least the first 18,457,734,525,360,901,453,873,570 decimal digits of e!

    The trick is that you need to take advantage of the fact that e=\lim_{n\rightarrow\infty}(1+\frac{1}{n})^n. We just have to construct a very very large value of n two times. We will use n=9^{4^{7\times 6}}=3^{2^{85}}. This gives us:
    e\sim (1+9^{0-4^{7\times 6}})^{3^{2^{85}}}.

    I did not discover this approximation. You can see it, and other pandigital approximations, here.

  2. (Not really the question you asked, but I felt like the closest pandigital might actually be in the range of computer brute-force, then spent an hour trying to get a reasonably tight upper bound on the number of cases one would have to examine (composition of binary operations is like building a binary tree with our operations on the internal nodes and our numbers at the leaves; 4862 unlabeled binary trees with ten leaves, times 10! permutations of numbers, times 6^9 ways to choose operations is “only” around 10^17, and we should be able to do better than that because addition and multiplication are commutative and presumably concatenation can only be applied to our starting digits), then gave up shortly after the point where I started to feel like 2^(18/5) really ought to be an integer.)

    • When I said estimate how close you can get, I was imagining that people might look at that 10^17 number, and try to estimate how many significant figures they might be able to get with that many randomly distributed numbers.

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