2002 AMC 12B Problem 15 — Solution | LIVE by Po-Shen Loh

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2002 AMC 12B Problem 15

Below is the professionally curated solution for Problem 15 of the 2002 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2002 AMC 12B solutions, or check the answer key.

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Concepts:place value digits

Difficulty rating: 1510

15.

How many four-digit numbers $N$ have the property that the three-digit number obtained by removing the leftmost digit is one ninth of $N?$

$4$

$5$

$6$

$7$

$8$

Solution:

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.

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