Putnam Problem!

$N = 10^{10^{10^n}} + 10^{10^n} + 10^n - 1.$ Write $n = 2^m k$ with $m$ a nonnegative integer and $k$ a positive odd integer.

For any nonnegative integer $j$ , $10^{2^m j} \equiv (-1)^j \pmod{10^{2^m} + 1}.$

Since $10^n \geq n \geq 2^m \geq m+1$ $\implies$ $10^n$ is divisible by $2^n$ and hence by $2^{m+1}$ , and similarly $10^{10^n}$ is divisible by $2^{10^n}$ and hence by $2^{m+1}$ .

It follows that $N \equiv 1 + 1 + (-1) + (-1) \equiv 0 \pmod{10^{2^m} + 1}.$

Since $N \geq 10^{10^n} > 10^n + 1 \geq 10^{2^m} + 1$ , we can conclude that $N$ is composite.