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ABCD is a 4 digit number where A,B,C and D are 4 distinctly separate digits between 1 and 9.

Using any of the following math operations and the digits A,B,C and D exactly once get the number ABCD.

Math operations allowed: + - x / ^ ! Square root and parenthesis. NO concatenation. There are at least 2 solutions. So answer with the use of the least number of math operations gets the tick.

Please no programming.

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    $\begingroup$ To be clear $(4-2)^{10}$ is not allowed? $\endgroup$
    – Daniel Mathias
    Aug 8, 2021 at 13:55
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    $\begingroup$ The 10 in the exponent is forbidden in capital letters, seems pretty clear. $\endgroup$
    – Bass
    Aug 8, 2021 at 14:07
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    $\begingroup$ @Bass is correct. Number 10 is not part of 1 to 9 digits. $\endgroup$
    – DrD
    Aug 8, 2021 at 14:13
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    $\begingroup$ @JaapScherphuis since we're looking for the solution with the least number of operators, it seems unlikely that the factorial and the square root are involved; we need to go from four separate numbers into only one, and unary operations don't change the number of numbers we have. $\endgroup$
    – Bass
    Aug 8, 2021 at 14:36
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    $\begingroup$ I voted to close this as underspecified to try to prevent it accumulating more and more answers which have different optimal solutions. $\endgroup$
    – bobble
    Aug 9, 2021 at 2:26

6 Answers 6

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On par with Bass, with preservation of digit order:

$5^5=3125\implies(3\times1+2)^5=3125$

Another, this time with reverse digit order:

$(6*1*9)^2=2916$

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    $\begingroup$ No, no, this is clearly better than mine, since it preserves the digit order. $\endgroup$
    – Bass
    Aug 8, 2021 at 14:27
  • $\begingroup$ Yes indeed @Daniel Mathias. And I just found a solution with factorials involved. $\endgroup$
    – DrD
    Aug 8, 2021 at 14:32
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In order and very brief

$4096 = 4^{0^9+6}$

... and also illegal since it contains a zero. Thanks @Bass.

If we are allowed parentheses to mean binomial coefficients:

$6435 = \begin{pmatrix}\begin{pmatrix} 6 \\ 4 \end{pmatrix} \\ 3+5 \end{pmatrix}$

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    $\begingroup$ This is so beautiful OP should immediately change the rules to lift the ban on zeroes. $\endgroup$
    – Bass
    Aug 8, 2021 at 17:17
  • $\begingroup$ I agree, zero is not evil. $\endgroup$
    – Daniel Mathias
    Aug 8, 2021 at 17:49
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Unless parens count as an operation, this should be optimal:

$(\frac{8}{2}-1)^7 = 2187$

(Of course I'll be keeping an eye out for a solution without parens, and with the digits in the correct order, but that might take a while.)

Here's one without parens, so this is definitely minimal in terms of operations:

$6\times4^5+1 = 6145$

Doesn't have the digit order style points though. After trying pretty much all of the likely candidates, and many unlikely ones too, I'm going to call it: I don't think there is a three operation solution without parens that has the digits in the correct order.

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  • $\begingroup$ Definitely one of the two solutions I had @Bass. I would be interested in knowing how you got there. $\endgroup$
    – DrD
    Aug 8, 2021 at 14:20
  • $\begingroup$ @DrD looked for any x^y with four distinct digits ABCD, then tried to make x and y out of A,B,C and D. Seemed easiest to do any maths while still in single digits. $\endgroup$
    – Bass
    Aug 8, 2021 at 14:24
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    $\begingroup$ You can get the first in order with a minor modification (2+1^8)^7. $\endgroup$
    – loopy walt
    Aug 8, 2021 at 17:23
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No fancy ordering.

$1296 = 6^{\frac{9-1}{2}}$

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How about:

$ 8192 = 2^{{\sqrt{9}}! - 1 + 8} $

Nobody has been using the monadic operations, so I did.

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    $\begingroup$ $8^{1+\sqrt 9} \times 2$ is in order and uses fewer operations. $\endgroup$
    – loopy walt
    Aug 9, 2021 at 15:24
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using factorial (!) is a great help to make big numbers using a simple digit num... Based on this, we can find the following answers:

In correct order:

$2163 = (2-1+6!)*3 $

$5167 = 5! + 1 + 6 + 7!$

$4368 = (\sqrt{4}*3)* (6!+8)$

No ordering:

$3625 = (6!+5)*(3+2)$

$5761 =7!+6!+1^5$

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  • $\begingroup$ These do use more math operations then the minimum, so they are not optimal, and optimal solutions have already been posted $\endgroup$
    – bobble
    Aug 13, 2021 at 13:56

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