hkjkstah Posted July 19, 2016 Share Posted July 19, 2016 Only using: ! ; ( ; ) ; / and digits: 4 ; 5 Can you make an equations to get result = 2p/s: we have45/5 = 94!=245/4=1.25 Find more quiz like this here! Challenge your mind with the HARDEST QUIZ in MATH ad removed Link to comment Share on other sites More sharing options...
uncool Posted July 19, 2016 Share Posted July 19, 2016 (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 1 Link to comment Share on other sites More sharing options...
John Cuthber Posted July 19, 2016 Share Posted July 19, 2016 Reminds me of a problem set by a maths teacher when I was at school. https://en.wikipedia.org/wiki/Four_fours 1 Link to comment Share on other sites More sharing options...
hkjkstah Posted July 20, 2016 Author Share Posted July 20, 2016 (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 Link to comment Share on other sites More sharing options...
uncool Posted July 21, 2016 Share Posted July 21, 2016 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? 1 Link to comment Share on other sites More sharing options...
hkjkstah Posted July 22, 2016 Author Share Posted July 22, 2016 so excellent My friends answer is the same with your answer (44/4)!/(45/5)!/55What about using: (;!;/ and digits: 2;3 to get result =41 ? Link to comment Share on other sites More sharing options...
uncool Posted July 22, 2016 Share Posted July 22, 2016 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)! Link to comment Share on other sites More sharing options...
hkjkstah Posted July 22, 2016 Author Share Posted July 22, 2016 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! Link to comment Share on other sites More sharing options...
uncool Posted July 22, 2016 Share Posted July 22, 2016 (edited) 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 Link to comment Share on other sites More sharing options...
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