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#1 2012-01-31 16:05:35

juantheron
Member
Registered: 2011-10-19
Posts: 312

probability

If two natural no. x and y are choosen at random . If P is the probability that

is divisable by 5 and 7. the valu of P is

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#2 2012-02-01 07:53:57

TheDude
Member
Registered: 2007-10-23
Posts: 361

Re: probability

For a number to be divisible by 5 its last digit must be 0 or 5.  Make a list of the final digits of the squares of all single digit numbers:

0 ->  0 -> 0
1 ->  1 -> 1
2 ->  4 -> 4
3 ->  9 -> 9
4 -> 16 -> 6
5 -> 25 -> 5
6 -> 36 -> 6
7 -> 49 -> 9
8 -> 64 -> 4
9 -> 81 -> 1

This means that the probability of the last digit of the square of a number being 0 is 10%, 1 is 20%, 4 is 20%, 5 is 10%, 6 is 20%, and 9 is 20%.  You can pair 0's and 5's with other 0's and 5's, 1's with 9's and 4's, 4's with 1's and 6's, 6's with 4's and 9's, and 9's with 1's and 6's.  This means that the probability of x^2 + y^2 being divisible by 5 is (0.2*0.2) + (0.2*0.4) + (0.2*0.4) + (0.2*0.4) + (0.2*0.4) = 0.36 = 36%.

For x^2 + y^2 to be divisible by 7, both x and y must be divisible by 7 (I think, I don't have a proof for this).  That means that there is a 1/7 * 1/7 = 1/49 chance that x^2 + y^2 is divisible by 7.  Multiply that by 36% to get 1/49 * 9/25 = 9/1225.


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#3 2012-02-01 08:21:40

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: probability

Hi TheDude;

Theorem: All squares are of the form 7k, 7k+1, 7k+2, 7k+4.

Supposing one of the squares, let's say x^2 wlog is not divisible by 7. So it must be of the form 7k+1, 7k+2, 7k+4.

It is easy to see that 7k+1 cannot be added to any of the above 3 to make a number divisible by 7.

For instance:

7k+1 + 7k+1 = 14k+2 (not divisible by 7)
7k+1 + 7k+2 = 14k+3 (not divisible by 7)
7k+1 + 7k+4 = 14k+5 (not divisible by 7)

We can exhaust all the 6 cases in the same way. So both numbers must be divisible by 7.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#4 2012-02-01 08:27:14

TheDude
Member
Registered: 2007-10-23
Posts: 361

Re: probability

Interesting.  How do you know that all squares are of the form 7k, 7k+1, 7k+2, 7k+4?  It's not immediately obvious to me why this is true.


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#5 2012-02-01 10:14:17

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: probability

Use the cyclic nature of modulo 7 on the squares.

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.
.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#6 2012-02-02 05:47:45

juantheron
Member
Registered: 2011-10-19
Posts: 312

Re: probability

Thanks Dude, Bobbym

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#7 2012-07-13 08:19:58

zetafunc.
Guest

Re: probability

bobbym wrote:

Use the cyclic nature of modulo 7 on the squares.

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.
.

Hi,

Is there a similar property for the cubes, or for the nth power?

#8 2012-07-13 08:21:37

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: probability

Hi;

Of course. Try them for yourself!


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#9 2012-07-13 08:22:47

zetafunc.
Guest

Re: probability

Okay, I will keep trying. I tried mod 7, 11, 13, 17 and 19 on the cubes and didn't notice any pattern though...

#10 2012-07-13 08:26:38

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: probability

Hi;

Cubes mod 7.

{1,1,6,1,6,6,0,1,1,6,1,6,6,0,1,1,6,1,6,6,0...}


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#11 2012-07-13 08:28:44

zetafunc.
Guest

Re: probability

Oh, I noticed that pattern, but I was wondering whether or not there was a cyclic pattern for mod n for the cubes.

#12 2012-07-13 08:30:00

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: probability

Same pattern for all mod n? No. Each mod is different. They can not be the same.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#13 2012-07-13 08:31:38

zetafunc.
Guest

Re: probability

Okay... so are the squares the only powers that exhibit a cyclic nature for mod n (n = 7)?

#14 2012-07-13 08:35:47

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: probability

All mod n's will have a cyclic nature. They are a remainder, what is left over after a division is done. They go from 0... n-1.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#15 2012-07-13 10:07:49

anonimnystefy
Real Member
From: Harlan's World
Registered: 2011-05-23
Posts: 16,049

Re: probability

Hi zeta

All powers show cyclic nature mod 7.


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