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Author Topic: A small mathmatical problem.  (Read 4315 times)
Unity
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« on: August 01, 2009, 07:11:12 AM »

somebody sent me this simple mathmatical problem, can you solve them?  two of them have been solved as an example you need to do the rest of it.  in any case the total must remain 6, you are allowed to use any mathmatical operation, but not allowed to add any new number:

1 1 1 = 6
2 + 2 + 2 = 6 ----->example
3 3 3 = 6
4 4 4 = 6
5 5 5 = 6
6 + 6 - 6 = 6 ----->example
7 7 7 = 6
8 8 8 = 6
9 9 9 = 6
« Last Edit: August 03, 2009, 02:38:44 AM by Unity » Logged
Khurasanzad
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« Reply #1 on: August 02, 2009, 03:40:11 PM »

Have you also the answer of it? Without adding new numbers you can´t work with it. Maybe if you use algorithm but it´s difficult.

3x3-3=6

« Last Edit: August 03, 2009, 01:43:01 AM by Unity » Logged
Unity
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« Reply #2 on: August 03, 2009, 01:43:21 AM »

Have you also the answer of it? Without adding new numbers you can´t work with it. Maybe if you use algorithm but it´s difficult.

3x3-3=6



The one you have solved is correct.  And yes, it is possible to do it without adding new numbers.  Try again and the others too.  Keep in mind that all mathmatical operations are allowed including division, subraction, addition and multiplication.
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Unity
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« Reply #3 on: August 03, 2009, 02:53:55 AM »

Have you also the answer of it? Without adding new numbers you can´t work with it. Maybe if you use algorithm but it´s difficult.

3x3-3=6



You dont have to go that high, the logarithm is not needed to solve it.  if you want to use logaritm you need to have the table, if you go that way it will be a bit complicated.
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Khurasanzad
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« Reply #4 on: August 03, 2009, 06:05:07 AM »

Have you also the answer of it? Without adding new numbers you can´t work with it. Maybe if you use algorithm but it´s difficult.

3x3-3=6



You dont have to go that high, the logarithm is not needed to solve it.  if you want to use logaritm you need to have the table, if you go that way it will be a bit complicated.

Is it allowed to use exponentiation? I ask you because It´s not really adding new numbers but somehow it is...for ex. 4*3=64. 3 is here the exponent. Have you already solved it?
« Last Edit: August 03, 2009, 09:43:41 AM by Khurasanzad » Logged
PORS
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« Reply #5 on: August 04, 2009, 04:21:18 PM »

Here is for 1, 4, 5, 6 and 7:

(1+1+1)!=6

(4+4)-(root of 4)=6

5+(5/5)=6

(6-6)+6=6

7-(7/7)=6

By the way, check with Wolfram website to see if it's correct. Wolfram is a great source of knowledge. USE IT!

http://www.wolframalpha.com/

- Pors
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Unity
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« Reply #6 on: August 06, 2009, 11:34:32 AM »

Here is for 1, 4, 5, 6 and 7:

(1+1+1)!=6

(4+4)-(root of 4)=6

5+(5/5)=6

(6-6)+6=6

7-(7/7)=6

By the way, check with Wolfram website to see if it's correct. Wolfram is a great source of knowledge. USE IT!

http://www.wolframalpha.com/

- Pors

yes pors bro, they are all correct, but 8 and 9 was missing.
« Last Edit: August 06, 2009, 01:15:09 PM by Unity » Logged
Unity
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« Reply #7 on: August 06, 2009, 01:14:32 PM »

Have you also the answer of it? Without adding new numbers you can´t work with it. Maybe if you use algorithm but it´s difficult.

3x3-3=6



You dont have to go that high, the logarithm is not needed to solve it.  if you want to use logaritm you need to have the table, if you go that way it will be a bit complicated.

Is it allowed to use exponentiation? I ask you because It´s not really adding new numbers but somehow it is...for ex. 4*3=64. 3 is here the exponent. Have you already solved it?
yes i have solved them, you are allowed to use any mathmatical method.  i give you a hint, exponentials are not needed to solve it.
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