15. How many 3-digit numbers are there with the property that the 2-digit number obtained by deleting the middle digit is equal to one ninth of the original 3-digit number?
A 1
B 2
C 3
D 4
E 5
15. D Let the three digits be $a, b, c$. Then the value of the 3-digit number is $100 a+10 b+c$ and the 2-digit number (made by deleting the middle digit) is $10 a+c$.
Hence $100 a+10 b+c=9(10 a+c)$, which gives $10(a+b)=8 c$, or $5(a+b)=4 c$. Hence $c$ is divisible by 5 . But $c$ cannot be zero (because this would mean $a+b=0$, and the digits cannot all be zero). So $c=5$. Then $a+b=4$, giving four possibilities: $a=1, b=3$ or $a=2, b=2$ or $a=3, b=1$ or $a=4, b=0$. These all give valid solutions $(135=9 \times 15,225=9 \times 25,315=9 \times 35,405=9 \times 45)$.