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This is a variation on the formation of numbers theme.

Given $1, 2, 3, x$ in order, and the operations plus, minus, times, divide, and brackets, what single digits can x NOT be?

i.e. find a solution for $x$ from $0$ to $9$, or prove there is no solution.

For example if $x=4$ we have:

$ (1+2)\times3-4=5$

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2 Answers 2

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It is possible for x = {0-9}.

$X=0: \;\;1 \times 2 + 3 + 0 = 5$
$X=1: \;\;1 \times 2 \times3 - 1 = 5$
$X=2: \;\;1 + 2 \times 3 - 2 = 5$
$X=3: \;\;1 - 2 + 3 + 3 = 5$
$X=4: \;\;(1 + 2) \times 3 - 4 = 5$
$X=5: \;\;(1 + 2) / 3 \times 5 = 5$
$X=6: \;\;1 + 2 / 3 \times 6 = 5$
$X=7: \;\;1 / 2 \times (3 + 7) = 5$
$X=8: \;\;(1 - 2) \times 3 + 8 = 5$
$X=9: \;\;1 - 2 - 3 + 9 = 5$

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  • $\begingroup$ they have to be in order $\endgroup$
    – JMP
    Mar 9, 2019 at 5:12
  • $\begingroup$ and done! Here you go! $\endgroup$
    – Cheri
    Mar 9, 2019 at 5:20
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    $\begingroup$ I don't think negation is allowed. A subtraction (minus) operation requires two operands. $\endgroup$
    – Amorydai
    Mar 9, 2019 at 5:29
  • $\begingroup$ Not sure if it was required, but answer has since been fixed to remove negation. $\endgroup$
    – Cheri
    Mar 9, 2019 at 5:53
  • $\begingroup$ You don't need the brackets in lines X=1 and X=2. $\endgroup$
    – Arnaud Mortier
    Mar 9, 2019 at 10:50
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Other solutions:

$X=1:\;\;1+2+3-1=5$
$X=1:\;\;(1\times2+3)\times1=5$
$X=1:\;\;1\times2+3\times1=5$

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