1. Last digit of (xyz)n(xyz)n is the same as that of znzn;
2. Determine the cyclicity number cc of zz;
3. Find the remainder rr when nn divided by the cyclisity;
4. When r>0r>0, then last digit of (xyz)n(xyz)n is the same as that of zrzr and when r=0r=0, then last digit of (xyz)n(xyz)n is the same as that of zczc, where cc is the cyclisity number.
• Integer ending with 0, 1, 5 or 6, in the integer power k>0, has the same last digit as the base.
• Integers ending with 2, 3, 7 and 8 have a cyclicity of 4.
• Integers ending with 4 (eg. (xy4)n(xy4)n) have a cyclisity of 2. When n is odd (xy4)n(xy4)n will end with 4 and when n is even (xy4)n(xy4)n will end with 6.
• Integers ending with 9 (eg. (xy9)n(xy9)n) have a cyclisity of 2. When n is odd (xy9)n(xy9)n will end with 9 and when n is even (xy9)n(xy9)n will end with 1.
Example: What is the last digit of 1273912739? Solution: Last digit of 1273912739 is the same as that of 739739. Now we should determine the cyclisity of 77:
1. 7^1=7 (last digit is 7)
2. 7^2=9 (last digit is 9)
3. 7^3=3 (last digit is 3)
4. 7^4=1 (last digit is 1)
5. 7^5=7 (last digit is 7 again!)
...
So, the cyclisity of 7 is 4.
Now divide 39 (power) by 4 (cyclisity), remainder is 3.So, the last digit of 1273912739 is the same as that of the last digit of 739739, is the same as that of the last digit of 7373, which is 33.