2002 AMC 12B Problems/Problem 15

Problem

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

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

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 $\mathrm{(D)}$ .

Solution 2

Since N is a four digit number, assume WLOG that $N = 1000a + 100b + 10c + d$ , where a is the thousands digit, b is the hundreds digit, c is the tens digit, and d is the ones digit. Then, $\frac{1}{9}N = 100b + 10c + d$ , so $N = 900b + 90c + 9d$ Set these equal to each other: $1000a + 100b + 10c + d = 900b + 90c + 9d$ $1000a = 800b + 80c + 8d$ $1000a = 8(100b + 10c + d)$ Notice that $100b + 10c + d = N - 1000a$ , thus: $1000a = 8(N - 1000a)$ $1000a = 8N - 8000a$ $9000a = 8N$ $N = 1125a$

Go back to our first equation, in which we set $N = 1000a + 100b + 10c + d$ , Then: $1125a = 1000a + 100b + 10c + d$ $125a = 100b + 10c + d$ The upper limit for the right hand side (RHS) is $999$ (when $b = 9$ , $c = 9$ , and $d = 9$ ). It's easy to prove that for an $a$ there is only one combination of $b, c,$ and $d$ that can make the equation equal. Just think about the RHS as a three digit number $bcd$ . There's one and only one way to create every three digit number with a certain combination of digits. Thus, we test for how many as are in the domain set by the RHS. Since $125\cdot7 = 875$ which is the largest $a$ value, then $a$ can be $1$ through $7$ , giving us the answer of $\boxed {D) 7}$

IronicNinja~

2002 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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2002 AMC 12B Problems/Problem 15

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Problem

Solution

Solution 2

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