Strayer Elementary Number Theory 1. 5. 84.

So I found afterwards that if you let b=200...0 (with k 0's), the proof still works but gives you exactly k 1's, not at least k 1's like in the proof below. Here's a QRD:

You still have (a,b)=d=1 because d|a and d|b implies d|a and d|18b. So

d|a-18b=200...0-(2)(9)(11...1)=200...0-(2)(99...9)=200...0-(2)(100...0-1)=200...0-200...0+2=2. But clearly d!=2 because d|a=11...1, which is odd. So d=1.

So Thm.12.1 still applies, telling us there are infinitely primes ending in at least k 1's. But since b=200...0, then the leading non-zero digits of nb will actually be an even integer. So the leading non-zero digits of nb will never end with a 1, just a 0,2,4,6, or 8. So the only 1's in the trailing part of the number will be the k 1's found in a. Ergo each of a,a+b,a+2b,...,a+nb,... ends in exactly k 1's.