Challenge - Keypad puzzle

[Albert Chan]

(02-13-2020 09:09 PM)pinkman Wrote:  

(02-13-2020 08:15 PM)Albert Chan Wrote:  gcd(2365, 2563) = 11
gcd(5698, 5896) = 22

You’re almost there! We’re looking for a prime common factor.

From above 1 case, the common prime factor is 11.

To prove, say, "counter-clockwise" number is divisible by 11
Modulo 11, for least significant digits, say x, at the 4 corners:

1. x + (x+1)*10 + (x-2)*100 + (x-3)*1000 ≡ x - (x+1) + (x-2) - (x-3) ≡ 0
2. x + (x-3)*10 + (x-4)*100 + (x-1)*1000 ≡ x - (x-3) + (x-4) - (x-1) ≡ 0
3. x + (x-1)*10 + (x+2)*100 + (x+3)*1000 ≡ x - (x-1) + (x+2) - (x+3) ≡ 0
4. x + (x+3)*10 + (x+4)*100 + (x+1)*1000 ≡ x - (x+3) + (x+4) - (x+1) ≡ 0
   

For the "clockwise" numbers, just imagine the weight of 1,10,100,1000 are reversed.
Modulo 11, we still get the RHS result, since 100≡1, 1000≡10≡-1