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

Latest revision as of 00:37, 22 November 2024

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 1

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~

Solution 2

Let Let $a$ denote the leftmost digit of $N$ and let $x$ denote the three-digit number obtained by removing $a$ . Then $N=1000a+x=9x$ and it follows that $1000a=8x$ . Dividing both sides by 8 yields $125a=x$ . All the values of $a$ in the range 1 to 7 result in three-digit numbers, hence there are $\boxed{D)7}$ values for $N$ .

∼Glowworm

@@ Line 5: / Line 5: @@
 \qquad\mathrm{(B)}\ 5
 \qquad\mathrm{(C)}\ 6
-\qquad\mathrm{
+\qquad\mathrm{(D)}\ 7
-\mathrm{(E)}\ 8</math>
+\qquad\mathrm{(E)}\ 8</math>
 == Solution ==
@@ Line 12: / Line 12: @@
-== Solution 2   ==
+== Solution 1   ==
 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>
@@ Line 34: / Line 34: @@
 IronicNinja~
+==Solution 2==
+Let Let <math>a</math> denote the leftmost digit of <math>N</math> and let <math>x</math> denote the three-digit number obtained by removing <math>a</math>. Then <math>N=1000a+x=9x</math> and it follows that <math>1000a=8x</math>. Dividing both sides by 8 yields <math>125a=x</math>. All the values of <math>a</math> in the range 1 to 7 result in three-digit numbers, hence there are <math>\boxed{D)7}</math> values for <math>N</math>.
+∼Glowworm
 == 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

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