17. How many three-digit numbers are increased by 99 when their digits are reversed?
A 4
B 8
C 10
D 80
E 90
17. D The value of ' $a b c$ ' equals $100 a+10 b+c$. We need to find integers of the form ' $a b c$ ' such that ' $c b a$ ' $=$ ' $a b c$ ' +99 .
So $100 c+10 b+a=100 a+10 b+c+99$, that is $99 c=99 a+99$. This condition simplifies to $c=a+1$. So there are 10 such integers of the form ' $1 b 2$ ', 10 of the form ' $2 b 3$ ', 10 of the form ' $3 b 4$ ', $\ldots$, and 10 of the form ' $8 b 9$ '. Therefore the required number is $8 \times 10=80$.