thanks a lot!

I tried to do something similar,

Briliant :) smartes evever . Everyone try iterate and caluclate and there is the betyfull bitwise

<< is bitwise shift left; that's basically multiplying by the powers of 2:

• << 1 == * 2**1 == * 2
• << 8 == * 2**8 == * 256
• etc.

Similarly, >> is bitwise shift right; which is (floor)dividing by the powers of 2:

• >> 1 == // 2**1 == // 2
• >> 8 == // 2**8 == // 256
• etc.

I like the simplicity of the solution. It's a rare case where using "magic numbers" and manually writing out each of the possibilities (instead of a loop) is actually okay and still quite readable.

One small suggestion: consider using bitwise | instead of +=. I think the result would be just as readable.

I tried to find where in the kata is the issue you are trying to mark here, but failed. Every mention of (2^200) ^ (2^300) results in 6 as the last digit. Can you please clarify what needs to be corrected? Or if there is no mistake anywhere in the kata, please check this issue as resolved.

Update: Okay, I think I found and corrected it (it was in Python-specific examples). Please mark this as resolved if satisfied.