The hardest quiz in math :) Can you solve it?

[hkjkstah]

Only using: ! ; ( ; ) ; / and digits: 4 ; 5
Can you make an equations to get result = 2

p/s: we have
45/5 = 9
4!=24
5/4=1.25

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[uncool]

(4!/(((((4/((4/4)/4))!/((((((4!/4)/(4/(4!/4)))!)/((4!/4)/(4/(4!/4))))/4)/((4!/4)!))!)/4)/4)/5))/4

Fun fact: this can be done just using 4s.

 

(4!/(((((4/((4/4)/4))!/((((((4!/4)/(4/(4!/4)))!)/((4!/4)/(4/(4!/4))))/4)/((4!/4)!))!)/4)/4)/((((4!/4)!)/(4!/4))/4!)))/4

[John Cuthber]

Reminds me of a problem set by a maths teacher when I was at school.

https://en.wikipedia.org/wiki/Four_fours

[hkjkstah]

(4!/(((((4/((4/4)/4))!/((((((4!/4)/(4/(4!/4)))!)/((4!/4)/(4/(4!/4))))/4)/((4!/4)!))!)/4)/4)/5))/4

Fun fact: this can be done just using 4s.

 

(4!/(((((4/((4/4)/4))!/((((((4!/4)/(4/(4!/4)))!)/((4!/4)/(4/(4!/4))))/4)/((4!/4)!))!)/4)/4)/((((4!/4)!)/(4!/4))/4!)))/4

so amazing answer

but my friends only use 18 characters to achieve this

[uncool]

If you're using 4 and 5 as digits, not just numbers, 20 characters:

 

((55/5)!/(45/5)!)/55

 

I'm sure that with some manipulation of parentheses, this could be gotten down to 18 characters, but I'm curious; what's your solution?

[hkjkstah]

so excellent
My friends answer is the same with your answer (44/4)!/(45/5)!/55

What about using: (;!;/ and digits: 2;3 to get result =41 ?

[uncool]

A slightly harder problem:

 

Prove that using just (, ), /, !, and the number 2 (note: not the digit) that it is possible to get any positive integer.

 

As for that one, my immediate thought is:

 

((((((3/((3/3)/3))!)/((3!)!))/3!)/2)!/(((((3/((3/3)/3))!)/((3!)!))/3!)/2))/((((3!)!)/3!)/3)!

[hkjkstah]

1=2/2 2=2 3= (2/(2/2/2))!/2!/(2/(2/2/2)) = 4!/2!/4

assume that we can get any positive integer from 1->n

 

now:
if n%2==1 =>exist k<=n: k/(2/2/2)=k*2= n+1=>done

if n%2==0 =>exist k<=n k/(2/2/2)=k*2= n+2 => n+1 =(n+2)!/(n+2)/n!

[uncool]

46 characters:

 

(((332/2/2)!/(332/2/2))/((3/((3/3)/3/3/3))!)/2

 

Also: well done, yup, that's my solution too.

Edited July 23, 2016 by uncool