6
$\begingroup$

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$

$\endgroup$
11
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ they have to be in order $\endgroup$ – JonMark Perry Mar 9 at 5:12
  • $\begingroup$ and done! Here you go! $\endgroup$ – Arch2K Mar 9 at 5:20
  • 1
    $\begingroup$ I don't think negation is allowed. A subtraction (minus) operation requires two operands. $\endgroup$ – Amorydai Mar 9 at 5:29
  • $\begingroup$ Not sure if it was required, but answer has since been fixed to remove negation. $\endgroup$ – Arch2K Mar 9 at 5:53
  • $\begingroup$ You don't need the brackets in lines X=1 and X=2. $\endgroup$ – Arnaud Mortier Mar 9 at 10:50
0
$\begingroup$

Other solutions:

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.