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Assemble a formula using the numbers $5$, $0$, $1$, and $7$ in any order to make 89. You may use the operations $x + y$, $x - y$, $x \times y$, $x \div y$, $x!$, $\sqrt{x}$, $\sqrt[\leftroot{-2}\uproot{2}x]{y}$ and $x^y$, as long as all operands are either $7$, $0$, $1$, or $5$. Operands may of course also be derived from calculations e.g. $10+(7*5)$. You may also use brackets to clarify order of operations, and you may concatenate two or more of the four digits you start with (such as $7$ and $5$ to make the number $75$) if you wish. You may only use each of the starting digits once and you must use all four of them. I'm afraid that concatenation of numbers from calculations is not permitted, but answers with concatenations which get $89$ will get plus one from me.

Double, triple, etc. factorials (n-druple-factorials), such as $5!! = 5 \times 3 \times 1$ are not allowed, but factorials of factorials are fine, such as $((5-1)!)! = 24!$. I will upvote answers with double, triple and n-druple-factorials which get $89$, but will not mark them as correct.

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    $\begingroup$ Just to be clear: it isn't valid to make 68 and then rotate the number, right? $\endgroup$ – Racso Sep 29 '18 at 22:10
  • $\begingroup$ @Racso - I love a solution like that and would upvote the answer if you post it, but I'm afraid that it is not the solution $\endgroup$ – tom Sep 29 '18 at 22:53
  • $\begingroup$ So just to be clear here, all operands have to be the single-digit numbers 0, 1, 5, or 7? Or is combining digits to form multi-digit numbers allowed? $\endgroup$ – gparyani Sep 30 '18 at 2:16
  • $\begingroup$ "and you may concatenate two or more of the four digits you start with (such as 7 and 5 to make the number 75) if you wish" $\endgroup$ – Racso Sep 30 '18 at 3:03
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    $\begingroup$ Can we use negative signs? $\endgroup$ – ubadub Sep 30 '18 at 6:51
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Here is lateral thinking answer:

$(5-0!)\times17=68$

then

up side down it is $89$.

Here is the actual answer:

$\sqrt{\frac{\sqrt{5!+1}!}{7!}+0!}$

How did I find?

First of all, I believed that the actual result I am looking for should be $89^2=7921$ and since we have 0 and 1 in our hands, I suspected that it should be just around +1 -1 from $89^2$ and noticed that $7920$ has lots of divisors and we have factorial option too. I noticed that $7921-1=11\times10\times9\times8=\frac{11!}{7!}$, so we have $7!$ to eliminate after $8$ but we need to find 11 with 5 and 0, or 5 and 1. Then 5! is $120$, which is just 1 value away from $11^2$ and the rest was easy and quick.

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  • $\begingroup$ Great job :-) well done $\endgroup$ – tom Sep 30 '18 at 11:50
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    $\begingroup$ I don't see the usefulness of these challenges, as it is kind of trivial to write a software that can bruteforce the answer $\endgroup$ – StefanS Sep 30 '18 at 17:41
  • $\begingroup$ @StefanS you may not able to write a code to find the answer with many operation options, it could take forever or long enough to cancel the run, otherwise everbody would find it in an instant. $\endgroup$ – Oray Sep 30 '18 at 17:44
  • $\begingroup$ Totally agree, but these are challenges, that have a certain degree of complexity. $\endgroup$ – StefanS Sep 30 '18 at 17:45
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Using double factorial, but no concatenation:

$5!!\times(7-1)-0!$

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    $\begingroup$ nice double factorial answer -- plus one $\endgroup$ – tom Sep 29 '18 at 20:59
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Something very close (an error of .61) is

$$ \sqrt\frac{5^7}{10} = 88.3883 $$

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  • $\begingroup$ Amazing that answer gets so close - plus one $\endgroup$ – tom Sep 30 '18 at 10:06
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Here's a lateral thinking attempt-

$17\times5+0!=86$ flip the $6$ to make a $9$ and we get $89$

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  • $\begingroup$ Nice answer +1, but sorry not the solution $\endgroup$ – tom Sep 30 '18 at 10:05
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An answer using double factorials:

$7!!-15-0!=89$

I'll try to get the real answer too.

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    $\begingroup$ Nice answer with double factorials .. plus one... $\endgroup$ – tom Sep 29 '18 at 20:35
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I believe

$71- (\sqrt{5-0!}) $

is 69.

Yay.

FUN ANSWER "Rearranging the digits"

EDIT

Oops, I misread it. I'll keep working on it. 5! is only 31 more than 89. The answer is tantalizingly close.

Hey, can we use any of the round, floor, ceiling, or truncate functions?

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  • $\begingroup$ Where is the zero? Don't forget to include 0!. $\endgroup$ – Brendon Shaw Sep 29 '18 at 20:00
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    $\begingroup$ The answer is supposed to be 89. This makes 69. $\endgroup$ – Weather Vane Sep 29 '18 at 20:00
  • $\begingroup$ oops, i fixed it. $\endgroup$ – Alto Sep 29 '18 at 20:01
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    $\begingroup$ I am sorry, I am too naive to understand why a wrong answer gets upvotes. It should be a fun comment. $\endgroup$ – Weather Vane Sep 29 '18 at 20:04
  • $\begingroup$ @tom So is this correct? $\endgroup$ – PotatoLatte Sep 29 '18 at 23:01
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I have something close, with an error of $0.006$.

$\sqrt{\sqrt{(5! + 0!)} \times (7 - 1)!} = \sqrt{11 \times 6!} = 88.994$

It is close but not close enough. A correct answer was given following similar lines, so I have posted as far as I got.

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  • $\begingroup$ Very close, plus one $\endgroup$ – tom Sep 30 '18 at 11:51
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My attempt

If we are allowed to use fibonacci. $f(7 + 5 +(1-0!))$

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  • $\begingroup$ Square isn’t a valid operator. $\endgroup$ – Quintec Sep 30 '18 at 0:50
  • $\begingroup$ Sorry my mistake $\endgroup$ – Chief VOLDEMORT Sep 30 '18 at 0:56
  • $\begingroup$ (You should delete this before it gathers downvotes, since it’s not a valid answer) $\endgroup$ – Quintec Sep 30 '18 at 1:09
  • $\begingroup$ 'Another attempt' looks pretty ingenious, but as suspected the fibonacci is not allowed... and I also suggest you might delete the first attempt $\endgroup$ – tom Sep 30 '18 at 10:08
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I use APL, in 8 chars... Need to attach a .jpg. First char is "ceiling" (round up), and the "*" is exponent.

( "ceiling" 71 ** .05 ) 89

Screen capture of sAPL session The sAPL interpreter can be downloaded for free (no adverts or tracking) from Google Play Store, and you can check the answer on any Android phone. Look for "GEMESYS" or "sAPL".

APL is great. :)

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  • $\begingroup$ Oops! I just read the rules of the puzzle, and I think the use of the ceiling operator does not qualify. Mea culpa! Sorry! :) $\endgroup$ – Mark Langdon Oct 1 '18 at 23:51

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