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andrezito [222]
4 years ago
10

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

Mathematics
1 answer:
Phoenix [80]4 years ago
6 0

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.

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