The remainder is zero. In fact 10^n + 11^n is divisible by 21 whenever n is an odd number.

You can prove this by showing that when n is odd, (a + b) is a factor of (a^n + b^n).

You can factorise (a^n + b^n) as:

(a + b)(a^(n-1) – a^(n-2)b + a^(n-3)b^2 - … - ab^(n-2) + b^(n-1))

When you multiply the massive second bracket by (a + b) almost everything cancels out, leaving only (a^n + b^n). Here is a specific example using n=5:

(a+b)(a^4 – a^3.b + a^2.b^2 – a.b^3 + b^4)

Multiplying each term in the second bracket first by a:

(a^5 – a^4.b + a^3.b^2 – a^2.b^3 + a.b^4)

And then by b:

(a^4.b – a^3.b^2 + a^2.b^3 – a.b^4 + b^5)

Adding those two brackets leaves only the first and last terms:

(a^5 + b^5)

So as long as a, b and n and integers, the whole second bracket will be too. Therefore (10 + 11) divides into (10^101 + 11^101).