Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2002 AMC 12B Problems/Problem 15"
Line 11: Line 11:
 
Let <math>N = \overline{abcd} = 1000a + \overline{bcd}</math>, such that <math>\frac{N}{9} = \overline{bcd}</math>. Then <math>1000a + \overline{bcd} = 9\overline{bcd} \Longrightarrow 125a = \overline{bcd}</math>. Since <math>100 \le \overline{bcd} < 1000</math>, from <math>a = 1, \ldots, 7</math> we have <math>7</math> three-digit solutions, and the answer is  <math>\mathrm{(D)}</math>.  
 
Let <math>N = \overline{abcd} = 1000a + \overline{bcd}</math>, such that <math>\frac{N}{9} = \overline{bcd}</math>. Then <math>1000a + \overline{bcd} = 9\overline{bcd} \Longrightarrow 125a = \overline{bcd}</math>. Since <math>100 \le \overline{bcd} < 1000</math>, from <math>a = 1, \ldots, 7</math> we have <math>7</math> three-digit solutions, and the answer is  <math>\mathrm{(D)}</math>.  
  
 +
 +
== Solution 2 - Longer but might make more sense ==
 +
Since N is a four digit number, assume WLOG that <math>N = 1000a + 100b + 10c + d</math>, where a is the thousands digit, b is the hundreds digit, c is the tens digit, and d is the ones digit.
 +
Then, <math>\frac{1}{9}N = 100b + 10c + d</math>, so <math>N = 900b + 90c + 9d</math>
 +
Set these equal to each other:
 +
<cmath>1000a + 100b + 10c + d = 900b + 90c + 9d</cmath>
 +
<cmath>1000a = 800b + 80c + 8d</cmath>
 +
<cmath>1000a = 8(100b + 10c + d)</cmath>
 +
Notice that <math>100b + 10c + d = N - 1000a</math>, thus:
 +
<cmath>1000a = 8(N - 1000a)</cmath>
 +
<cmath>1000a = 8N - 8000a</cmath>
 +
<cmath>9000a = 8N</cmath>
 +
<cmath>N = 1125a</cmath>
 +
 +
Go back to our first equation, in which we set <math>N = 1000a + 100b + 10c + d</math>,
 +
Then:
 +
<cmath>1125a = 1000a + 100b + 10c + d</cmath>
 +
<cmath>125a = 100b + 10c + d</cmath>
 +
The upper limit for the right hand side (RHS) is <math>999</math> (when <math>b = 9</math>, <math>c = 9</math>, and <math>d = 9</math>).
 +
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 <math>125\cdot7 = 975</math> which is the largest a value, then a can be <math>1</math> through <math>7</math>, giving us the answer of <math>\boxed {D) 7}</math>
 +
 +
IronicNinja~
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2002|ab=B|num-b=14|num-a=16}}
 
{{AMC12 box|year=2002|ab=B|num-b=14|num-a=16}}

Revision as of 20:46, 29 October 2018

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 - Longer but might make more sense

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 $125\cdot7 = 975$ which is the largest a value, then a can be $1$ through $7$, giving us the answer of $\boxed {D) 7}$

IronicNinja~

See also

2002 AMC 12B (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12B_Problems/Problem_15&oldid=98416"