Determine(with proof) the last digit of 2
2n for n ≥ 2.
The first three powers of 2 end in 2, 4, 8, 6. The next one returns to 2, and that begins the next cycle of four members.
2^n where n>1 will always be a multiple of 4, so no matter how high 2^n gets, the exponent will be a multiple of 4, and you'll be at the fourth member of the cycle: 6.
|
|
Posted by Charlie
on 2024-11-17 10:51:38 |
solution (spoiler)
