2002 AMC 12B Problem 15

Let $a$ be the leading digit and $x$ the three-digit number after removing it, so $N=1000a+x.$ The condition $N=9x$ gives $1000a=8x,$ i.e. $x=125a.$ For $a=1,\ldots,7$ this makes $x$ a three-digit number, while $a=8$ gives $x=1000.$ So there are $7$ such numbers. Thus, the correct answer is D.