probability

juantheron · 2012-01-31 16:05:35

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

TheDude · 2012-02-01 07:53:57

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.

bobbym · 2012-02-01 08:21:40

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.

TheDude · 2012-02-01 08:27:14

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.

bobbym · 2012-02-01 10:14:17

Use the cyclic nature of modulo 7 on the squares.

.
.
.

juantheron · 2012-02-02 05:47:45

Thanks Dude, Bobbym

zetafunc. · 2012-07-13 08:19:58

bobbym wrote:

Use the cyclic nature of modulo 7 on the squares.
.
.
.

Hi,

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

bobbym · 2012-07-13 08:21:37

Hi;

Of course. Try them for yourself!

zetafunc. · 2012-07-13 08:22:47

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

bobbym · 2012-07-13 08:26:38

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...}

zetafunc. · 2012-07-13 08:28:44

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

bobbym · 2012-07-13 08:30:00

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

zetafunc. · 2012-07-13 08:31:38

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

bobbym · 2012-07-13 08:35:47

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.

anonimnystefy · 2012-07-13 10:07:49

Hi zeta

All powers show cyclic nature mod 7.

Math Is Fun Forum

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

probability

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

Re: probability

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

Re: probability

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

Re: probability

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

Re: probability

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

Re: probability

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

Re: probability

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

Re: probability

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

Re: probability

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

Re: probability

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

Re: probability

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

Re: probability

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

Re: probability

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

Re: probability

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

Re: probability

Board footer