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

(s)
 
(Problem)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
How many four-digit numbers <math>N</math> have the property that the three-digit number obtained by removing the leftmost digit is one night of <math>N</math>?
+
How many four-digit numbers <math>N</math> have the property that the three-digit number obtained by removing the leftmost digit is one ninth of <math>N</math>?
  
 
<math>\mathrm{(A)}\ 4
 
<math>\mathrm{(A)}\ 4

Revision as of 21:30, 24 February 2008

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 $\mathrm{(D)}$.

See also

2002 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions