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

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

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?

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...

futb0l
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 $x \neq 0$

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

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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