2021/103) Show that there are infinite numbers of the form $10^n+3$ that are composite

Because the number is not divsible by 2,3,5 for any n so let us check if is divisible by 7 for some n. 

Now as 7 is co-prime to 10 so as per fermats little theorem 

$10^6 \equiv 1 \pmod 7\cdots(1)$ 

By checking from1 to 6 we see that 

$3^4 = 81 \equiv  4 \pmod 7$

or  $10^4 = 81 \equiv  4 \pmod 7$ as $10\equiv 3 \pmod 7$

using (1) we get 

$10^{6k+ 4} = \equiv  4 \pmod 7$

or  $10^{6k+ 4} + 3 = \equiv  0 \pmod 7$

hence divisible by 7 and hence composite

so there are infinite numbers as k goes from 1 onwards are composite for n = 6k + 4