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i assume we cannot consider 0

Humph...

Note that any number must be 7k + r with r = 0,1,...6

(7k + r)^3 = (7k + r)^2
would that help - i'm clutching at straws to be honest!!

So I think it is 7k+1

"(7k + r)^3 = (7k + r)^2"
Maybe expand these instead of setting them equal...

So, I think u mean to say 7k+1

I mean every number can be expressed as 7k + r with r = 0,1,...6
Yes, I said that above....

oh.....THAT WAS A HINT

The question does say 7k or 7k +1

@NewbieCarrot Considering certain remainders is a good way to go....

(7k+r)^2 = 49k^2 + 14kr + r^2 = 7(7k^2+2kr) + r^2

(7k+r)^3 = 343k^3 +147k^2r +21kr^2 +r^3 =
7(49k^3 +21k^2r +3kr^2) +r^3

Not done yet....

7(7k^2+2kr) + r^2
7(49k^3 +21k^2r +3kr^2) +r^3
have the same form with 7k+r while r=r^2=r^3 (mod 7)

OK, the reaminders are the same, and.....?

The only possible answer is r=0 or 1. And we are done

n = 7k+r
n^2 = 7x + r^2
n^3 = 7y + r^3

r = [1,6]

0 also

Yes, that's right, well done.

Just include r= 0 as well....

oh yes... .

Is there this rule:
If n=x MOD y then, n^a = x^a MOD y

Is there this rule:
If n=x MOD y then, n^a = x^a MOD y
Yes (r>=1)

(a>=1)

GREAT....... Then I think I proved if n=1,2,3,4,5,6 MOD 7 Then n^6=1 MOD 7

a^(p-1) = 1 mod p (gcd a,p = 1)

p prime

a^(p-1) = 1 mod p (p prime, gcd a,p = 1) (OR a^p = a mod p)
Fermat's Little Theorem

Indirectly related to the "necklace2 problem....

WHAT!

But how???? That is related to permutation and combination.

(n^p-n)/2p + [n(p+1/2) -n]/2 + n

Did u mean that every colour has only one bead.