Elementary math that professors cant solve

[killerinstinct]

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]

9+\frac{9}{9}=20_{(base 5)}

 

[Grizzlycomet]

Why shouldn't professors solve what even I can solve? \frac{(9+9)}{.9}=20

 

[Zurtex]

9+\frac{9}{9}=20_{(base 5)}

lol

9*9 - 9 = 20_{(base 36)}

9*9 + 9 = 20_{(base 45)}

9*(9 + 9) = 20_{(base 81)}

9! + 9 - 9 = 20_{(base 181440)}

9! + 9 + 9 = 20_{(base 181449)}

9^9 - 9 = 20_{(base 193710240)}

9^9 + 9 = 20_{(base 193710249)}

etc...

 

[killerinstinct]

Bases are not ELEMENTARY MATH!

 

[arildno]

Bases are not ELEMENTARY MATH!

I think they are elementary

 

[ahrkron]

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]

Okay then, I cheated, I'm terribly sorry.

 

[fourier jr]

Bases are not ELEMENTARY MATH!

My textbook called Elementary # Theory has bases in it...

 

[Simon666]

( 9² * sqrt(9) ) - sqrt(9) = (81*3)-3 = 240. I choose to use the division symbol / to cross the 4 et voila.

 

[HallsofIvy]

In any case, why did you say professor's can't solve this? How many professors did you try?

 

[killerinstinct]

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]

Do you know about the "four fours" variation of this theme?

 

[killerinstinct]

Explain questions? (using four 4s to equal something)?

 

[arildno]

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]

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

 

[Bob3141592]

Okay then, I cheated, I'm terribly sorry.

Very naughty. As punishmnt, you should be whipped with a bundle of rays.

 

[Grizzlycomet]

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

How about 4+4+\frac{4}{\sqrt{4}}

 

[Euphoriet]

http://mathforum.org/ruth/four4s.puzzle.html

http://www.jimloy.com/puzz/4-4s.htm

 

[Gunni]

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:

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.

 

[Njorl]

How about 4+4+\frac{4}{\sqrt{4}}

I think using a square root implicity requires a "2".

 

[dellianaL]

professors can't solve this? wow.

 

[mikesvenson]

How about 4+4+\frac{4}{\sqrt{4}}

that works for me

 

[StonedPanda]

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

 

[Simon666]

Sweet. Good work!

 

[Njorl]

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

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...(8-8^1/3)x8⁰
9 9 9 = 6...9-9/(9^1/2)

Njorl

 

[killerinstinct]

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

 

[mikesvenson]

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...(8-8^1/3)x8⁰
9 9 9 = 6...9-9/(9^1/2)

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..

 

[plover]

Get -1 using 0,0,0

0 + 0 - 0!

Or to get 6:

(0! + 0! + 0!)!

 

[futb0l]

0 - 0^0
is this qualified?

 

[futb0l]

<br /> -\cos{0} - 0 + 0<br />

and mm...

<br /> - ( \sin ^2 0 + \cos ^2 0 ) + 0<br />

 

[Zurtex]

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]

Or for those whose tastes run to notation sadism:

- \lim_{0\rightarrow0} 0!​

 

[futb0l]

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]

read that: http://mathforum.org/dr.math/faq/faq.0.to.0.power.html

 

[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

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:

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]

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]

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]

9+\frac{9}{9}=20_{(base 5)}

I rarely do math for fun so my so I don't 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?