# Elementary math that professors cant solve

#### Njorl

StonedPanda said:

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

the square root and the square kind of mess it up, but it's still pretty damn sweet
That equals 20.

sqrt(9)=3
9-sqrt(9)=6
6!=720

6^2=36

720/36=20

Njorl

#### Njorl

1 1 1 = 6................(1+1+1)!
2 2 2 = 6................2+2+2
3 3 3 = 6................3x3-3
4 4 4 = 6................(4!/4)x40
5 5 5 = 6................5+5/5
6 6 6 = 6................6+6-6
7 7 7 = 6................7-7/7
8 8 8 = 6................(8-81/3)x80
9 9 9 = 6................9-9/(91/2)

Njorl

#### killerinstinct

How come I never thought of those answers that Njorl did. Hmmm, something is wrong with my brain. lol!

#### mikesvenson

Njorl said:
1 1 1 = 6................(1+1+1)!
2 2 2 = 6................2+2+2
3 3 3 = 6................3x3-3
4 4 4 = 6................(4!/4)x40
5 5 5 = 6................5+5/5
6 6 6 = 6................6+6-6
7 7 7 = 6................7-7/7
8 8 8 = 6................(8-81/3)x80
9 9 9 = 6................9-9/(91/2)

Njorl
what does 8^0 mean?
what does is equal?

and where do you get 9-9/9^1/2
doesnt 9-9=0, then 0/9^1/2=0?

#### Muzza

x^0 is equal to 1 for all real (and complex) x.

9 - 9/9^(1/2) is interpreted as $$9 - \frac{9}{9^{1/2}}$$. If he had meant $$\frac{9 - 9}{9^{1/2}}$$, he would have written (9 - 9)/9^(1/2). The parantheses are important :P

#### ExecNight

How bout this..

Get -1 using 0,0,0 Oh and this is mathematically possible without using any tricks..

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#### plover

Homework Helper
ExecNight said:
Get -1 using 0,0,0
0 + 0 - 0!

Or to get 6:

(0! + 0! + 0!)!

#### futb0l

$$0 - 0^0$$
is this qualified?

#### futb0l

$$-\cos{0} - 0 + 0$$

and mm...

$$- ( \sin ^2 0 + \cos ^2 0 ) + 0$$

#### Zurtex

Homework Helper
futb0l said:
$$0 - 0^0$$
is this qualified?
Strictly speaking $0^0$ is not defined. As:

$$x^0 = \left( x^1 \right) \left( x^{-1} \right)$$

Therefore:

$$x^0 = \frac{x}{x}$$

Which means $x^0 = 1$ when $x \neq 0$

#### plover

Homework Helper
Or for those whose tastes run to notation sadism:

$$- \lim_{0\rightarrow0} 0!$$​

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#### futb0l

Zurtex said:
Strictly speaking $0^0$ is not defined. As:

$$x^0 = \left( x^1 \right) \left( x^{-1} \right)$$

Therefore:

$$x^0 = \frac{x}{x}$$

Which means $x^0 = 1$ when $x \neq 0$
mmm.. if you do 0^1 in google, it will come up as 1.
and ... http://mathforum.org/dr.math/faq/faq.0.to.0.power.html

so i don't think
$x^0 = 1$ when $x \neq 0$
is true.

#### futb0l

there should be a rule that says when any number is to the power of 0 it will be equal to 1.

#### ExecNight

Most of em are true solutions..

Now the funny thing here is we are getting something from noting..

How come we can get 1 from 0 by using only 0? That always makes my head iching...

#### NoNose

Hi
First: Sorry if my english isn´t correct or couldn´t be understand, but i´m trying to.

Without to resume the discussion, if factorial and bases are elemental math, i´m thinking i´ve found a solution for the problem down this text for all positive and negative real numbers and the 0:
$$\left( \left( x^2 \right) ^0 + \left( x^2 \right) ^0 + \left( x^2 \right) ^0 \right) ! =6$$

Gunni said:
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 ):
6 + 6 - 6 = 6

Have fun.

#### Zurtex

Homework Helper
NoNose said:
Hi
First: Sorry if my english isn´t correct or couldn´t be understand, but i´m trying to.

Without to resume the discussion, if factorial and bases are elemental math, i´m thinking i´ve found a solution for the problem down this text for all positive and negative real numbers and the 0:
$$\left( \left( x^2 \right) ^0 + \left( x^2 \right) ^0 + \left( x^2 \right) ^0 \right) ! =6$$
I like that , but using the square function is kind of using a 2 really (where as the square root actually has a symbol). So perhaps before anyone complains about this it could be easily fixed as:

$$\left( |x|^0 + |x|^0 + |x|^0 \right) ! = 6$$

For $x \neq 0$

#### Bartholomew

Njorl said:
10 (44-4)/4

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

Njorl
4 * 4 - 4! / 4 works

#### physicsuser

arildno said:
$$9+\frac{9}{9}=20_{(base 5)}$$
I rarely do math for fun so my so I dont know much math indepth. From what I know base 2 or binary numbers are like this

01 this means that 0*(2^1)+1*(2^0)=1

So in base 5 it would be

0,1,2,3,4 = 0*(5^4)+1*(5^3)+2*(5^2)+3*(5^1)+4*(5^0)=194

how do you get 9 in base 5? Is it a different base system or something?

edit--------------

Oh I think I know

9 =14
14/14=1
14+1=20
right?

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