Question

14. Prove that the number 10^2019 - 1 is composite

Answer 1

By the binomial theorem,

$10^{2019}=(9+1)^{2019}=\displaystyle\sum_{n=0}^{2019}\binom{2019}n9^{2019-n}$

The last term in the sum, when $n=2019$, is

$\dbinom{2019}{2019}9^{2019-2019}=1$

which is eliminated, leaving us with

$10^{2019}-1=\displaystyle\sum_{n=0}^{2018}\binom{2019}n9^{2019-n}$

$10^{2019}-1=\displaystyle\binom{2019}09^{2019}+\binom{2019}19^{2018}+\cdots+\binom{2019}{2018}9$

Each term in the remaining sum has a common factor of 9, so $10^{2019}-1$ must be composite.