Using the rules in question 57701, make exactly 113.

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Request for Question Clarification by blazius-ga on 23 Aug 2002 03:49 PDT

According to http://astronomy.swin.edu.au/~pbourke/fun/4444/ , the two
known solutions to this puzzle requires some non-standard notation:

4!/.(sqrt(4))` + sqrt(4)/.4      
to be read as
4!/.222222... + sqrt(4)/.4 = 108 + 5 = 113

or

(.4 - .4`%) / .4`% + 4!
to be read as
(.4 - .00444444...) / .00444444... + 4! = 0,39555555.../.00444444... +
24 = 89 + 24 = 113

If any of these solutions are acceptable, please tell me.

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Clarification of Question by michael2-ga on 23 Aug 2002 04:19 PDT

No, 'fraid I'm not going to accept either of those as proper answers. 
Once we allow made-up non-standard notations, it gets a bit too easy!

Hi micheal,

   Good day and thanks for your question. I just love solving these
type of puzzles. In fact, I did a similar one (but easier one) today

Coming to your answer, 113 = 4!/.(sqrt(4))` + sqrt(4)/.4

' indicated recurring digit

4!=24
.sqrt(4)'=0.2222222222222222.............
4!/.sqrt(4)' = 108
2/.4 = 5
108+5=113

There you go :-)

Coming to what I did today, make the number 24 using four zeros which
is simple
(0!+0!+0!+0!)!

Have a good day and thank you for using this service

Cheers,
aditya2k

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Clarification of Answer by aditya2k-ga on 23 Aug 2002 05:24 PDT

Oops, didn't see your clarification that this answer is not
allowed...I'll send a message to the editors to remove this answerq

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Request for Answer Clarification by michael2-ga on 23 Aug 2002 05:42 PDT

OK.  Should I do anything at this end - eg reject the answer & re-post
- or can you just get the answer removed as a mistake?

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Clarification of Answer by aditya2k-ga on 23 Aug 2002 05:53 PDT

Hi micheal,

   You need not do anything at your end, for I have found the solution!
         ____________  |
        /  __________  |^4
113=   /  /       .    |
     \/ \/ [4^4*(.4)]  |

In simpler words

sqrt(sqrt(grtintfunc((4^4)*recur(.4))))^4

4^4=256
64*.44444444... = 113.777777777777777777777777777664
[113.777777777777777777777777777664]=113
sqrt(sqrt(113^4)) = 113

I hope this is acceptable to you

Cheers,
aditya2k

OK, yes I'll accept the answer clarification.  Clever use of the
integer function, represented by square brackets.  Why didn't I think
of that?

(4! + 4)*4 + sqrt(sqrt(4))
=28*4 + 1
=113

tne
sqrt(sqrt(4)) = sqrt(2) which is certainly not = 1
so that one doesnt work