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In this puzzle you need to obtain the highest score with the following rules

1) choose two digits from 1, 2, 3, 4, 5, 6, 7, 8 and 9. These digits are represented by two letters, let us say $T$ and $M$. You are also given the digit 0.

2) make up an equation using all three of your digits, $T$, 0 and $M$, once. The numerical result of the equation should be a three digit number which can be $TM0$, $T0M$, $MT0$, $M0T$, $0MT$ and $0TM$. You may not concatenate digits, but you may use as many as the following symbols as you would like $+$, $-$, $\div$, $\times$, !, $\sqrt{}$, (, ). Note that multiple ! symbols are considered each to be factorial functions. See the note below.

3) Your score is your three digit number divided by the sum of your digits. So for example if you start with 0, 2, 4 then your equation could be

$$(0*2)+4!=024$$

which would score $024 \div (2+4+0) = 24\div 6 = 4$


This question was inspired by this question

Note that: $4!! = (4!)! = 24!$ this is allowed. Double, triple etc. factorials are not allowed; $4!! \ne 4 \times 2$

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  • 1
    $\begingroup$ Glad to help you in a modest way $\endgroup$ – Uvc May 29 at 19:02
  • $\begingroup$ Is the use of the decimal point allowed? $\endgroup$ – hexomino May 29 at 23:29
  • $\begingroup$ @hexomino no, but i will of course upvote a creative answer using one.... $\endgroup$ – tom May 30 at 3:17
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    $\begingroup$ I'm voting to close this question because it seems too open-ended to have a unique best answer. $\endgroup$ – Rand al'Thor May 30 at 17:20
  • $\begingroup$ @Randal'Thor I appreciate your point, but It seems to me that the answer given is fine and explains things well - sorry if I have taken too long to accept the answer. $\endgroup$ – tom May 30 at 22:38
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I'm going to try:

$(7-2+0!)!=6!=720$.
Score 720/9=80.

This I believe is the best available, as the only combo's that could beat it are:

$410,\;510,\;610,\;710,\;810,\;910,\;820$ and $920$

and I haven't managed to make any of these.

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