5
$\begingroup$

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}.

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

$\endgroup$
6
  • $\begingroup$ they have to be in order $\endgroup$
    – JMP
    Mar 9 '19 at 5:12
  • $\begingroup$ and done! Here you go! $\endgroup$
    – Cheri
    Mar 9 '19 at 5:20
  • 1
    $\begingroup$ I don't think negation is allowed. A subtraction (minus) operation requires two operands. $\endgroup$
    – Amorydai
    Mar 9 '19 at 5:29
  • $\begingroup$ Not sure if it was required, but answer has since been fixed to remove negation. $\endgroup$
    – Cheri
    Mar 9 '19 at 5:53
  • $\begingroup$ You don't need the brackets in lines X=1 and X=2. $\endgroup$ Mar 9 '19 at 10:50
1
$\begingroup$

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.