Use 6, 5 and 3 to make 57

Assemble a formula using the numbers $$5$$, $$6$$, and $$3$$ in any order to make $$57$$. 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 $$3$$, $$6$$, or $$5$$. Operands may of course also be derived from calculations e.g. $$3+(6*5)$$. 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 $$6$$ and $$5$$ to make the number $$65$$) if you wish. You may only use each of the starting digits once and you must use all three of them. I'm afraid that concatenation of numbers from calculations is not permitted, but answers with concatenations which get $$57$$ 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 $$((3)!)! = 6!$$. I will upvote answers with double, triple and n-druple-factorials which get $$57$$, but will not mark them as correct.

A finite number of operations should be used, but answers with an infinite number of operations will get +1 from me...

• Tom, you're posting a lot of these puzzles and they're all basically the same and -- excuse my frankness! -- becoming rather boring. Maybe give them a rest for a while? – Gareth McCaughan Sep 30 '18 at 17:28
• @GarethMcCaughan to be honest, some of them are very original, requires good logic deduction. but yes, would be good to give a rest for a while :) – Oray Sep 30 '18 at 17:31
• @GarethMcCaughan you could say the same about the large number of Riley riddles. Some people like one kind of puzzle, but not the other, etc. Isn't the time to stop when no-one answers, or there are a lot of very quick correct answers? – Weather Vane Sep 30 '18 at 17:34
• @GarethMcCaughan - point taken... I will retire from these riddles. – tom Sep 30 '18 at 17:50

$$5! - 63 = 57$$

• Great job well done – tom Sep 30 '18 at 17:47

hexomino beat me, but I have another:

$$\sqrt{5\times6!}-3=57$$

• IMO this answer without number concatenation is better. – Weather Vane Sep 30 '18 at 17:40
• @WeatherVane, but both are within the rules, so the sightly earlier answer gets the solution, but both get plus one. – tom Sep 30 '18 at 17:48

Can I have:

$$56+\sqrt{\sqrt{\dots{3}}}$$

• That's a dotty answer. Interesting but is it in the rules? – Weather Vane Sep 30 '18 at 17:39
• @WeatherVane it counts as an infinite number of operations, so gets plus one. – tom Sep 30 '18 at 17:48
• @tom I see you changed the rules... before this answer was posted. – Weather Vane Sep 30 '18 at 17:54
• @WeatherVane - yes it was an edit a couple of minutes after posting because I figured out this solution, but my screen says this answer was 25 minutes ago and the edit to the question was 33 minutes ago --- that said Jon may well have not seen the edit when he posted the answer. -- Anyway its a good answer and it has my plus one :-) – tom Sep 30 '18 at 18:00
• @tom your rules were edited about 8 minutes before this question was posted ... although you say it will get an upvote, so I deleted my previous comment. NB I like your number puzzles. – Weather Vane Sep 30 '18 at 18:26

Is it possibly...

(5*6*3)-(5*6+3)

• Nice answer, but you have used each digit twice instead of once – tom Oct 1 '18 at 18:22

Consider $$f = 6 + 6 + 6 + 6 + 5 * 6 + 3$$. This equals $$57$$

• Note in the problem statement: "You may only use each of the starting digits once..." – Nick A Sep 30 '18 at 22:43

5! - 63 gives 120 - 63 = 57

I was just looking at possible combinations with factorials.

• Same as the accepted answer! – jmarkmurphy Oct 1 '18 at 16:26