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.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

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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