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Nice quizz for the new year


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#1 Napoleon 1er

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Posted 29 December 2015 - 01:07 PM

Hi all mathiacs,

 

One of my friend gave me a good quizz for math lovers:

with the following numbers 1;3;4 and 6, using each of them only once, with which calculation can you reach 24 as final result?

 

Have fun on stratego.com and happy new year to you all! :)


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#2 texaspete09

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Posted 29 December 2015 - 08:32 PM

(6^2/4-1)x3, but I don't know if we are supposed to square numbers.
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#3 sevenseas

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Posted 29 December 2015 - 08:48 PM

Another way:

 

6 * (4!/3!) * 1 =>

6 * 4 * 1 =>

 

24


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#4 sevenseas

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Posted 29 December 2015 - 08:49 PM

(6^2/4-1)x3, but I don't know if we are supposed to square numbers.

I don't think that counts because you use an extra 2 for the exponent


Edited by sevenseas, 29 December 2015 - 08:49 PM.

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#5 TheOptician

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Posted 29 December 2015 - 10:26 PM

(1^3)*6*4 - Again only if you are allowed 'to the power of'



#6 Napoleon 1er

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Posted 29 December 2015 - 10:28 PM

no 2, no !, no ^ ... there is a solution without all that!


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#7 sevenseas

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Posted 29 December 2015 - 11:41 PM

((arccos(1) * 3)) + (6 * 4)

 

:)


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#8 Astros 17

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Posted 29 December 2015 - 11:50 PM

log(1)*3+6*4


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#9 Astros 17

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Posted 29 December 2015 - 11:55 PM

arccos(sin(1)*3+6*4

 

Any function with 1 that returns 0 multiplied 3 + 6*4 works.

 

C(4,3)*6*1

 

det(6*4*x+31)

 

There are many many answers.


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#10 TheOptician

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Posted 29 December 2015 - 11:55 PM

6/3/1 = 2.  

 

Then put the 4 next to the 2.  

 

24.


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#11 TheOptician

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Posted 30 December 2015 - 01:10 AM

(14 - 6 ) * 3

#12 Napoleon 1er

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Posted 30 December 2015 - 08:16 AM

Lol ... there is one solution without function and without putting a number next to another ...
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#13 Astros 17

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Posted 30 December 2015 - 08:41 AM

6/(1-(3/4))


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#14 Napoleon 1er

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Posted 30 December 2015 - 08:47 AM

Congratulations­čÄć
If you don't know where you go ... you have a lot of chance to arrive elsewhere ...




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