where n is any positive integer. Since fractional powers of b have been eliminated in this way, and since a and b are both divisible by 5, we may conclude that d1n​+d2n​ is divisible by 10.
We now apply the above result twice, taking n=19 and n=82. In this way we obtain
d119​+d219​=10k1​ and d182​+d282​=10k2​,
where k1​ and k2​ are positive integers. Adding and rearranging these results gives
d119​+d182​=10k−(d219​+d282​),
where k=k1​+k2​. But d2​=15−220​=15+220​5​<31​.
Therefore, d219​+d282​<1. It follows that the units digit of 10k−(d219​+d282​) is 9.