21. Each of the positive integers $a, b$, and $c$ has three digits, and for each of these integers the first digit is the same as its last digit. Also $b=2 a+1$ and $c=2 b+1$. How many possibilities are there for the integer $a$ ?
A 0
B 1
C 2
D 3
E more than 3
21. C Let the digits of $a$ be 'pqp', so $a=101 p+10 q$. Also $c=2 b+1=2(2 a+1)+1=4 a+3$ which is $404 p+40 q+3$. This is less than 1000 , so $p=1$ or 2 .
If $p=2$, then $c=808+40 q+3$. This ends with digit $1(8+3=11)$ but $c$ is a 3-digit number greater than 808 so can't begin with 1 .
If $p=1$, then $c=404+40 q+3$. This ends with digit $4+3=7$, so must also begin with 7 , hence $700 \leq 404+40 q+3<800$ and thus $293 \leq 40 q<393$. Therefore $q=8$ or 9 .
When $q=8$, this gives $a=181, b=363, c=727$. When $q=9$, this gives $a=191, b=383$, $c=767$.
Hence there are two possibilities for $a, 181$ and 191 .