Difference between revisions of "2002 AMC 12B Problems/Problem 15"

Revision as of 16:37, 18 January 2008

Problem

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

$\mathrm{(A)}\ 4 \qquad\mathrm{(B)}\ 5 \qquad\mathrm{(C)}\ 6 \qquad\mathrm{(D)}\ 7 \qquad\mathrm{(E)}\ 8$

Solution

Let $N = \overline{abcd} = 1000a + \overline{bcd}$ , such that $\frac{N}{9} = \overline{bcd}$ . Then $1000a + \overline{bcd} = 9\overline{bcd} \Longrightarrow 125a = \overline{bcd}$ . Since $100 \le \overline{bcd} < 1000$ , from $a = 1, \ldots, 7$ we have $7$ three-digit solutions, and the answer is $\mathrm{(D)}$ .

2002 AMC 12B (Problems • Answer Key • Resources)
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Difference between revisions of "2002 AMC 12B Problems/Problem 15"

Revision as of 16:37, 18 January 2008

Problem

Solution

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