Melody

8/2=4     and    2+2=4

so tha answer is   4*4=16

Thanks guest, I have fixed mine now.  It was just a silly error on my part.

We are now in agreement.

Let the pieces be x, y and 5-(x+y)

where 

0

0

also

0<5-(x+y)<5

-5<-(x+y)<0

0

y< -x+5,     y>-x

Draw that up on a number plane and you get a triangle with an area of 12.5  that is your sample space.

Now graph the favourable area.   And work out what it is.

x>1

y>1

5-(x+y)>1

You can take over from here.

Quite possibly.  What do you base this assertion on?

Consider all positive integers between 3000 and 4000 which, when divided by any of the first three primes, leave a remainder of 1. 

Find the sum of all these integers.

(3000+x) mod 2 = 1     therefore  x is odd

(3000+x)mod 3=1         

so

(3000+x) mod6 =1

(3000+x)mod5 =1

so

(3000+x)mod30=1

The smallest number that meets this requirement is   x=1

the next is x=31,

then it goes up by 30.

The numbers will be     

3000(1+31+61+ ........ 991)

Sum=3000*34+(1+(1+30)+(1+2*30)+ ........ (1+33*30))  

         Consider the AP     a=1,  d=30,   n=34, 

\(\qquad S_{n}=\frac{n}{2}(a+L)\\ \qquad S_{34}=\frac{34}{2}(1+991)\\ \qquad S_{34}=17*(992)\\ \qquad S_{34}=16864\\\)

Sum= 3000*34+16864 =   118864    

I have just reapaired my formulas in red.  Now the answers to this interpretation of the question are the same.

Beats me.    Any takers for this one?

If you get a full answer can you explain it here please.

You can send me a private message to let me know you have added it.

Otherwise i might not notice.

Thanks :)

Thanks Chenxander,

I do not know either BUT

If rotations just mean the corner colours must stay in corners 

THEN

I get two different initial positons that can be rotated.

The green, which is my fixed one, can be in a corner, or in the middle of a side.  

If ithe green is in a corner then I get this

G 6 1

5 7 2

5 4 3 

7!*5 =   25200  permutations

If the green is in the middle of a side then I get this

6 G 5

4 7 3

3 2 1 

7!*3 = 15120 permutation

Now       25200+15120 = 40320                     

[Your given answer  of 20160 is exactly half of this which appears to me to be more than just a coincidence  ]   

According to my logic 40320 is the number of permutations if rotations of whole sides is considered the same.

However I have not considered reflections yet.There are 4 axes of symmetry so I want to divide by 2^4=16.

I do not know why the given answer is only divided by 2.

I personally do not know how to prove it but with all roots there is always 'that' number of roots,

so square root has 2 values

cube roots have 3 values

4th roots have 4 values  etc

I am sure if you google it you will find a proof.

Attn: chenxander

I am still waiting for your response.