Remainders

just simply think

15x6=90, remains 10.

15x66=990, remains 10.

15x666=9990, remains 10.

15x6666=99990, remains 10

so, whatever we extend(the power of 10), remainder will always be 10

$10^{\color{#D61F06}n} \equiv (\underbrace{999\cdots 999}_{\text{\# of } 9s = n-1}0 + 10) \equiv 10 \text{ (mod 15)}$ .

$\color{#D61F06}\text{Note:}$ For large enough natural number $n$ to solve this problem (see @Pi Han Goh 's remarks).

If $10$ mod $15 = 10$ , therefore $10^n$ mod $15 = 10$ ,

We can prove it by induction.

For $n=1$ , we know that $10$ mod $15= 10$ .

Suppose it is true for $n>0$ that $10^n$ mod $15 =10$ ,

then for $n+1$ , we have $10^{n+1}$ mod $15 = (10^n \times 10 )$ mod $15 =$

( $10^n$ mod $15 )(10$ mod $15) = (10)(10) = 10^2$ mod $15 = 10$ .

Thus, the above statement is true for $n>0$ .