mathematics - Concatenation divisible by 19

You are given a 9-digit number $a$. Let $b=a+1$.

Now, if you make a big number $N$ by concatenating $a$ and $b$, then $N$ is divisible by 19.

What is the number $a$?

Answer

For any $a<999{,}999{,}999$ we have $b\le 999{,}999{,}999$ which is 9 digits, too, so the $a$ part in the $N$ number is shifted by 9 decimal places. So $$N=10^9\cdot a+(a+1)=(10^9+1)a + 1 = 2,631,579\cdot 19\cdot a + 1$$ which is NOT divisible by $19$.

Let's try $a=999{,}999{,}999$ then. Hurray! We get $b=1{,}000{,}000{,}000$ and $N=9{,}999{,}999{,}991{,}000{,}000{,}000 = 526{,}315{,}789{,}000{,}000{,}000\cdot 19$

Hence $a=999{,}999{,}999.$