For any integer a show that a and a4n 1 have the same last

For any integer a, show that a and a^4n + 1 have the same last (decimal) digit.

Solution

We focus on only the last digit of a

Case 1. Last digit 1

Nothing to prove

Case 2. Last digit 2

2^4=16

Last digit of 2^4 is 6

Hence, Last digit of (2^4)^n is 6 because 6^m for any positive integer m has last integer 6

HEnce, 2^{4n+1} has last digit of :6*2=12 ie 2

Case 3: Last digit being 3

3^4=81

So last digit of 3^{4n} is 1

Hence last digit of 3^{4n+1} is 3

Case 4: 4^2 =16

Hence last digit of 4^{4n} has last digit 6

HEnce last digit of 4^{4n+1} is last digit of 6*4=24 ie 4

Case 5. Last digit 5

For all powers of 5 last digit is 5

Case 6. Last digit 6

For all powers of 6 last digit is 6

Case 7. Last digit 7

7^2 has last digit 9

HEnce ,7^4 has last digit same as last digit of 9^2=81 ie 1

HEnce, 7^{4n} has last digit =1

Hence, 7^{4n+1} has last digit 1*7=7

Case 8. Last digit 8

8^2 has last digit 4

HEnce,8^4 has last digit 6

Hence, 8^{4n} has last digit 6

Hence, 8^{4n+1} has last digit =same as last digit of 6*8=48 ie 8

9. Last digit 9

9^2=1

HEnce last digit of 9^{4n} is 1

HEnce last digit of 9^{4n+1} is 9

Hence proved