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.