Show that 27 x 23^n + 17 x 10^2n is divisible by 11 for all positive integers n. I know that modulos should aid in answering this problem,

27= 2 mode 5

23= 2 mode 1
17= 1 mode 6
10^2n= 100^n
100= 1 mode 9
So the problem reduces to
5*1^n+6*1^n
=5+6
=11

oops,typo error.

100=9 mode 1 the rest is correct.

To show that the expression 27 x 23^n + 17 x 10^2n is divisible by 11 for all positive integers n, we can use modular arithmetic.

First, let's consider the divisibility rule for 11. A number is divisible by 11 if the difference between the sum of its odd-placed digits and the sum of its even-placed digits is divisible by 11.

Now, let's express each term of the expression in terms of modulo 11.

27 x 23^n (mod 11):
Using the exponentiation rule, we can rewrite this as (27 mod 11) x (23 mod 11)^n (mod 11).

27 mod 11 is equivalent to 5, and (23 mod 11) is equivalent to 1.
So, the expression becomes 5 x 1^n (mod 11).

Since multiplying by 1 has no effect, we can simplify this to just 5 (mod 11).

17 x 10^2n (mod 11):
Using the exponentiation rule again, we can rewrite this as (17 mod 11) x (10^2 mod 11)^n (mod 11).

17 mod 11 is equivalent to 6, and (10^2 mod 11) is equivalent to 1.
So, the expression becomes 6 x 1^n (mod 11).

Again, multiplying by 1 has no effect, so we can simplify this to just 6 (mod 11).

Now, let's consider the entire expression:

(27 x 23^n) + (17 x 10^2n) (mod 11):
Using the distributive property, we can rewrite this as (27 x 23^n + 17 x 10^2n) (mod 11).

Substituting the values we found earlier:
(5 + 6) (mod 11).

Adding 5 and 6, we get 11 (mod 11), which is equivalent to 0.

Since the expression simplifies to 0 (mod 11), we can conclude that 27 x 23^n + 17 x 10^2n is divisible by 11 for all positive integers n.