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Can you find a way to make:

6 5 4 3 = 81

by concatenation and/or adding any of (and only) these mathematical operators:

  • +
  • -
  • ×
  • !
  • ÷
  • ^
  • standard parentheses ()

You cannot add other numbers to the equation, or re-order the existing numbers.
The result must be a mathematical equality.

Harder version...

Try to do it while only altering the left hand side!

Inspired by Make 5 5 5 5 = 19

Hopefully somewhat more challenging than my previous attempt (may be hard not to be!)

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    $\begingroup$ So $(6-5)\times 3^4=81$ won't work because I rearranged $3$ and $4$; and $6+\left(3\times 5^{\sqrt{4}}\right)=81$ won't work because there is at least a square root, right? $\endgroup$
    – Mr Pie
    Commented Aug 27, 2018 at 9:25
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    $\begingroup$ @user477343 correct on both accounts $\endgroup$
    – Jonathan Allan
    Commented Aug 27, 2018 at 9:39

11 Answers 11

Reset to default
37
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I thought a bit too much but I finally got it:

$\frac{6+5!*4}{3!}=81$

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  • $\begingroup$ Nice! Well done! $\endgroup$
    – R.D
    Commented Aug 27, 2018 at 8:00
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    $\begingroup$ Welcome to Puzzling! Very good. $\endgroup$
    – Jonathan Allan
    Commented Aug 27, 2018 at 8:02
  • $\begingroup$ Anyone looking for more to find, I have found another solution keeping the right hand side intact :) $\endgroup$
    – Jonathan Allan
    Commented Aug 27, 2018 at 9:45
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My solution: Just normal Math

$-6 + (5 + 4!) \times 3 = 81$

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    $\begingroup$ Welcome to Puzzling! Looks like the simplest solution to the harder version so far. $\endgroup$
    – Jonathan Allan
    Commented Aug 27, 2018 at 17:53
  • $\begingroup$ Why is this in the low-quality queue? It looks like a perfectly acceptable answer to me. Welcome to the site, btw! $\endgroup$
    – F1Krazy
    Commented Aug 27, 2018 at 18:19
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    $\begingroup$ @F1Krazy iirc low quality queue is automatic, based on length of answer(and probably being a new user factors in) $\endgroup$
    – Quintec
    Commented Aug 28, 2018 at 0:44
  • $\begingroup$ This is the simplest answer according to my computer solution. $\endgroup$
    – Neil G
    Commented Aug 28, 2018 at 12:47
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Using the notation of double factorial:

$$6!!+5!!+4!-3! = (6\times4\times2)+(5\times3\times1)+(4\times3\times2\times1)-(3\times2\times1)$$

WolframAlpha approves it.

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  • $\begingroup$ Nice! I did not see this one (maybe write out the way it evaluates since the double factorial operation is slightly lesser known (I see you linked to mathworld, but you know nice to have the working). $\endgroup$
    – Jonathan Allan
    Commented Aug 27, 2018 at 9:58
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    $\begingroup$ I've edited as your suggestion $\endgroup$
    – Anastasiya-Romanova 秀
    Commented Aug 27, 2018 at 10:07
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    $\begingroup$ Wow! That's the best one! $\endgroup$
    – kkm mistrusts SE
    Commented Aug 29, 2018 at 8:10
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My answer was (before the no swapping rule)

$(6+3) \times (5+4)= 9\times9=81$

Edit, after the no swapping rule

Step 1: $ 6+5+4+3=8+1$

and then

Step 2: $ 1+8=9$ (because in the previous statement lhs=18)

Finally, maintaining all the rules, this was also written before the harder version was mentioned):

$(6-5)+(4+3)=8\times1$

or

$(6-5)\times(4+3)=8-1$

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  • $\begingroup$ Hmm I thought "by concatenation and/or adding ..." was clear, must not be - I shall add "no reordering". Nice way to think outside the box though +1 $\endgroup$
    – Jonathan Allan
    Commented Aug 27, 2018 at 6:19
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    $\begingroup$ Oops XD. Will try again. Maybe $\endgroup$
    – R.D
    Commented Aug 27, 2018 at 6:20
  • $\begingroup$ @JonathanAllan Can I do it in two steps or do I have to do it in one step ? $\endgroup$
    – R.D
    Commented Aug 27, 2018 at 6:51
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    $\begingroup$ I don't understand. You can use parentheses... $\endgroup$
    – Jonathan Allan
    Commented Aug 27, 2018 at 6:59
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    $\begingroup$ @JonathanAllan the last one should be satisfactory :3. Should I remove the unnecessary answers? Or keep them as it is? $\endgroup$
    – R.D
    Commented Aug 27, 2018 at 7:08
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Here's a simple one (easy mode):

$$ 6\times(5-4)+3 = 8+1$$

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4
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My answer:

As concatenation allowed: $46 + $$35$$ =81$

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    $\begingroup$ +1 since you answered before my edit to disallow re-ordering :) $\endgroup$
    – Jonathan Allan
    Commented Aug 27, 2018 at 6:20
  • $\begingroup$ Yeah noticed that;) $\endgroup$
    – Preet
    Commented Aug 27, 2018 at 6:21
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    $\begingroup$ $(+1)$, but that rearranges order... $\endgroup$
    – Mr Pie
    Commented Aug 28, 2018 at 10:54
  • $\begingroup$ Thaks @user477343, actually OP edited the question after or when I was answering;) $\endgroup$
    – Preet
    Commented Aug 29, 2018 at 1:12
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Long time reader, first time answer-er:

By using some muscle to get the subtraction to be commutative, -(-(6 - (5-4)) - 3)=8^1

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3
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Something like:

$ 6 - 5 = 1 $ $ 4 + 3 = 7$

$1 + 7 = 8 * 1$

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    $\begingroup$ For the simple version of the challenge that indeed works (it's also been posted by R.D). You can do the steps using parentheses "(...)+(...)=...". By the way spoiler text is achieved by prefixing a line with >! (newlines can be placed inside these by adding two spaces to to end of the line and placing another >! on the next line (I see you are a regular on SO so I imagine you'll figure it all out easily - Welcome to your active-life at Puzzling!) $\endgroup$
    – Jonathan Allan
    Commented Aug 27, 2018 at 10:54
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    $\begingroup$ I'll take your tips in consideration next time.Thank you @jonathan $\endgroup$
    – Pedro Lobito
    Commented Aug 27, 2018 at 13:15
  • $\begingroup$ Also, write $\times$ to generate $\times$ for multiplication; you could also write $\ast$ to generate "$\ast$" with better formatting :) $\endgroup$
    – Mr Pie
    Commented Aug 28, 2018 at 10:55
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This feels like stretching the rules a little bit

65 + 4^(--3) = 81

This assumes that

the decrement operator -- (minus minus)

is allowed.

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    $\begingroup$ Wouldn't the operator in question need to come prior to the number? $\endgroup$
    – Taco
    Commented Aug 27, 2018 at 19:21
  • $\begingroup$ Since I have paranthesis, it can be on either side. $\endgroup$
    – infinitezero
    Commented Aug 27, 2018 at 20:52
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    $\begingroup$ I tested it with parentheses and it didn't do the operation until after execution as I expected. Is this just a software level thing, or is it the same in mathematics? $\endgroup$
    – Taco
    Commented Aug 27, 2018 at 20:58
  • $\begingroup$ I might have been wrong. I'm not familiar with this operator in mathematics, that's why I said this might be a little stretch. I'll change it to the front. $\endgroup$
    – infinitezero
    Commented Aug 27, 2018 at 22:25
  • $\begingroup$ In case someone wanted to stick to the programming style: 65 | 4 << --3 = 81 $\endgroup$
    – user27263
    Commented Aug 28, 2018 at 14:00
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By modifying both sides:

6-5+4+3 = 8 ÷ 1

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$654 + 3 = 81$, as long as you do the calculation in base 82.

Explanation:

$ 654 + 3 == 657 $ in base 10, subtract 1 gives 656, which is $ 8 * 82$

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  • $\begingroup$ This is really doing the calculation in base ten and then evaluating the result as if it were written in base eighty-two. Doing the calculation 654 + 3 in base eighty-two would be: forty-thousand-seven-hundred-and-fifty-eight plus three $\endgroup$
    – Jonathan Allan
    Commented Aug 28, 2018 at 18:47

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