Prove that 27(23^n)+17(10^2n)+22n is divisible by 11 for n belongs to the natural numbers

Proofs come in many shapes and sizes but this one is done via induction. When doing proof by induction, examiners love a list of a clear definite list of steps, the first being the base case, in this case it's n=1 (as n belongs to the natural numbers) so, 27(23¹)+17(10²¹)+221 = 27(23)+17(100)+22 = 621+1700+22 = 2343 (which when divided by 11 gives 213, a lovely whole number! therefore the base case has been proved true) Now that we've finished with with showing the base case true we can jump into the meat and bones of this proof. We're going to "Assume n=k true." so, "27(23^k)+17(10^2k)+22k = 11Z" where Z is a whole number. Next, let's consider the case of n=k+1 then, "27(23^k+1)+17(10^2(k+1))+22(k+1)" = "2723^k+2723¹+1710^2k+1710²+22k+221" Let's group up these terms, so "2723¹+1710²+221" is actually what we did in our base-case, it's 2343 (which is divisible by 11), next we have our "2723^k+1710^2k+22k" does this look familiar? It's because it's our assumption! So, by using our assumption we can see that "2723^k+1710^2k+22k" = 11Z and therefore, "2343+11Z=11(213+Z)", and for the finale the finishing sentence "We have shown that if it is true for n=k then it's true for n=k+1 since we've shown true for n=1 it must be true for all natural numbers"