# Make 6 5 4 3 = 81

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!)

• 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? – Mr Pie Aug 27 '18 at 9:25
• @user477343 correct on both accounts – Jonathan Allan Aug 27 '18 at 9:39

I thought a bit too much but I finally got it:

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

• Nice! Well done! – R.D Aug 27 '18 at 8:00
• Welcome to Puzzling! Very good. – Jonathan Allan Aug 27 '18 at 8:02
• Anyone looking for more to find, I have found another solution keeping the right hand side intact :) – Jonathan Allan Aug 27 '18 at 9:45

My solution: Just normal Math

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

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

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.

• 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). – Jonathan Allan Aug 27 '18 at 9:58
• I've edited as your suggestion – Anastasiya-Romanova 秀 Aug 27 '18 at 10:07
• Wow! That's the best one! – kkm Aug 29 '18 at 8:10

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

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

• 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 – Jonathan Allan Aug 27 '18 at 6:19
• Oops XD. Will try again. Maybe – R.D Aug 27 '18 at 6:20
• @JonathanAllan Can I do it in two steps or do I have to do it in one step ? – R.D Aug 27 '18 at 6:51
• I don't understand. You can use parentheses... – Jonathan Allan Aug 27 '18 at 6:59
• @JonathanAllan the last one should be satisfactory :3. Should I remove the unnecessary answers? Or keep them as it is? – R.D Aug 27 '18 at 7:08

Here's a simple one (easy mode):

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

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

• +1 since you answered before my edit to disallow re-ordering :) – Jonathan Allan Aug 27 '18 at 6:20
• Yeah noticed that;) – Preet Aug 27 '18 at 6:21
• $(+1)$, but that rearranges order... – Mr Pie Aug 28 '18 at 10:54
• Thaks @user477343, actually OP edited the question after or when I was answering;) – Preet Aug 29 '18 at 1:12

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

Something like:

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

$1 + 7 = 8 * 1$

• 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!) – Jonathan Allan Aug 27 '18 at 10:54
• I'll take your tips in consideration next time.Thank you @jonathan – CONvid19 Aug 27 '18 at 13:15
• Also, write $\times$ to generate $\times$ for multiplication; you could also write $\ast$ to generate "$\ast$" with better formatting :) – Mr Pie Aug 28 '18 at 10:55

This feels like stretching the rules a little bit

65 + 4^(--3) = 81

This assumes that

the decrement operator -- (minus minus)

is allowed.

• Wouldn't the operator in question need to come prior to the number? – Candleshark Aug 27 '18 at 19:21
• Since I have paranthesis, it can be on either side. – infinitezero Aug 27 '18 at 20:52
• 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? – Candleshark Aug 27 '18 at 20:58
• 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. – infinitezero Aug 27 '18 at 22:25
• In case someone wanted to stick to the programming style: 65 | 4 << --3 = 81 – user27263 Aug 28 '18 at 14:00

By modifying both sides:

6-5+4+3 = 8 ÷ 1

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

Explanation:

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

• 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 – Jonathan Allan Aug 28 '18 at 18:47