Elementary math that professors cant solve

Using only three 9's along with elementary math symbols like + or -, see if you can arrange them to represent the number 20. Remeber that 99/9=11.
 

arildno

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[tex]9+\frac{9}{9}=20_{(base 5)}[/tex]
 
Why shouldnt professors solve what even I can solve? [tex]\frac{(9+9)}{.9}=20[/tex]
 
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Zurtex

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arildno said:
[tex]9+\frac{9}{9}=20_{(base 5)}[/tex]
lol

[tex]9*9 - 9 = 20_{(base 36)}[/tex]

[tex]9*9 + 9 = 20_{(base 45)}[/tex]

[tex]9*(9 + 9) = 20_{(base 81)}[/tex]

[tex]9! + 9 - 9 = 20_{(base 181440)}[/tex]

[tex]9! + 9 + 9 = 20_{(base 181449)}[/tex]

[tex]9^9 - 9 = 20_{(base 193710240)}[/tex]

[tex]9^9 + 9 = 20_{(base 193710249)}[/tex]

etc...
 
Bases are not ELEMENTARY MATH!!!
 

arildno

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killerinstinct said:
Bases are not ELEMENTARY MATH!!!
I think they are elementary :biggrin:
 

ahrkron

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Regardless of how "elementary" bases are, the original problem says "the number 20" (which one can reasonably argue to be stated in base 10), instead of "a number with the representation '20' in some base".
 

arildno

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Okay then, I cheated, I'm terribly sorry.
 
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killerinstinct said:
Bases are not ELEMENTARY MATH!!!
My textbook called Elementary # Theory has bases in it.... :wink:
 
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( 9² * sqrt(9) ) - sqrt(9) = (81*3)-3 = 240. I choose to use the division symbol / to cross the 4 et voila. :devil:
 

HallsofIvy

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In any case, why did you say professor's can't solve this? How many professors did you try?
 
I know that there are many complicated (...) solutions to this problem. Many involving bases, but the most simplest solution is given by Grizzlycomet (look above) using only elementary basic math. It is not a matter of "professor not solving the problem", its just a FUN question! Don't interpet me wrong.
 

arildno

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Do you know about the "four fours" variation of this theme?
 
Explain questions? (using four 4s to equal something)?
 

arildno

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That's right; if I remember correctly, every number up to and including 12(?) can be written with 4 4's and standard math operations (no silly base shifts..)

I'm not absolutely sure about the last member of this set (i.e., 12), it's been a while since I saw it.

(Of course, lots of other numbers can be written using 4 fours too, but they are not consecutive..)
 

Njorl

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1 4x4/(4x4)
2 4x4/(4+4)
3 (4+4+4)/4
4 4+(4-4)/4
5 4+(4/4)^4
6 4+(4+4)/4
7 4+4-(4/4)
8 4+4x4/4
9 4+4+4/4
10 (44-4)/4
11 (4!+4)/4+4
12 (4!)x4/(4+4)


I had to use one "44". Is there a way to get 10 without resorting to this?

Njorl
 
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arildno said:
Okay then, I cheated, I'm terribly sorry.
Very naughty. As punishmnt, you should be whipped with a bundle of rays.
 
Last edited:
Njorl said:
I had to use one "44". Is there a way to get 10 without resorting to this?
Njorl
How about [tex]4+4+\frac{4}{\sqrt{4}}[/tex]
 
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There's another fun variation on this theme where you line up all the numbers from one to nine in threes and are supposed to make them add up to six by adding only plus, minus, division, multiplication, root and power signs (whole powers and roots, no logs!). You can also use ( and ) (forgot what they're called).

Like this:
Code:
1   1   1 = 6
2   2   2 = 6
3   3   3 = 6
4   4   4 = 6
5   5   5 = 6
6   6   6 = 6
7   7   7 = 6
8   8   8 = 6
9   9   9 = 6
For example (I hope I'm not ruining anything for anyone here :wink:):
6 + 6 - 6 = 6

Have fun.
 

Njorl

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Grizzlycomet said:
How about [tex]4+4+\frac{4}{\sqrt{4}}[/tex]
I think using a square root implicity requires a "2".
 
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professors can't solve this? wow.
 
Grizzlycomet said:
How about [tex]4+4+\frac{4}{\sqrt{4}}[/tex]
that works for me
 
How about this one?

[(9-sqrt(9))!]/[(sqrt(9)!)^2]

the square root and the square kind of mess it up, but it's still pretty damn sweet
 
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Sweet. Good work!
 

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