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 "2004 AMC 12B Problems/Problem 25"

m (→‎Solution)
(→‎Solution)
Line 7: Line 7:
  
 
Given <math>n</math> digits, there must be exactly one power of <math>2</math> with <math>n</math> digits such that the first digit is <math>1</math>. Thus <math>S</math> contains <math>603</math> elements with a first digit of <math>1</math>. For each number in the form of <math>2^k</math> such that its first digit is <math>1</math>, then <math>2^{k+1}</math> must either have a first digit of <math>2</math> or <math>3</math>, and <math>2^{k+2}</math> must have a first digit of <math>4,5,6,7</math>. Thus there are also <math>603</math> numbers with first digit <math>\{2,3\}</math> and <math>603</math> numbers with first digit <math>\{4,5,6,7\}</math>. By using [[complementary counting]], there are <math>2004 - 3 \times 603 = 195</math> elements of <math>S</math> with a first digit of <math>\{8,9\}</math>. Now, <math>2^k</math> has a first of <math>\{8,9\}</math> [[iff|if and only if]] the first digit of <math>2^{k-1}</math> is <math>4</math>, so there are <math>\boxed{195} \Rightarrow \mathrm{(B)}</math> elements of <math>S</math> with a first digit of <math>4</math>.
 
Given <math>n</math> digits, there must be exactly one power of <math>2</math> with <math>n</math> digits such that the first digit is <math>1</math>. Thus <math>S</math> contains <math>603</math> elements with a first digit of <math>1</math>. For each number in the form of <math>2^k</math> such that its first digit is <math>1</math>, then <math>2^{k+1}</math> must either have a first digit of <math>2</math> or <math>3</math>, and <math>2^{k+2}</math> must have a first digit of <math>4,5,6,7</math>. Thus there are also <math>603</math> numbers with first digit <math>\{2,3\}</math> and <math>603</math> numbers with first digit <math>\{4,5,6,7\}</math>. By using [[complementary counting]], there are <math>2004 - 3 \times 603 = 195</math> elements of <math>S</math> with a first digit of <math>\{8,9\}</math>. Now, <math>2^k</math> has a first of <math>\{8,9\}</math> [[iff|if and only if]] the first digit of <math>2^{k-1}</math> is <math>4</math>, so there are <math>\boxed{195} \Rightarrow \mathrm{(B)}</math> elements of <math>S</math> with a first digit of <math>4</math>.
 +
 +
== Alternate Solution ==
 +
 +
We can make the following chart for the possible loops of leading digits:
 +
<cmath>1 \rightarrow 2 \rightarrow 4 \rightarrow 8 \rightarrow 1</cmath>
 +
<cmath>1 \rightarrow 2 \rightarrow 4 \rightarrow 9 \rightarrow 1</cmath>
 +
<cmath>1 \rightarrow 2 \rightarrow 5 \rightarrow 1</cmath>
 +
<cmath>1 \rightarrow 3 \rightarrow 6 \rightarrow 1</cmath>
 +
<cmath>1 \rightarrow 3 \rightarrow 7 \rightarrow 1</cmath>
 +
 +
Let there be <math>x</math> of the sequences of <math>4</math> numbers, and let there be <math>y</math> of the sequences of <math>3</math> numbers. We note that a <math>4</math> appears only in the loops of <math>4</math>, and also we are given that <math>2^{2004}</math> has <math>604</math> digits.
 +
<cmath>3x+4y=2004</cmath>
 +
<cmath>x+y=603</cmath>
 +
Solving gives <math> x = 408</math> and <math>y = 195</math>, thus the answer is <math>(B)</math>.
  
 
== See also ==
 
== See also ==

Revision as of 18:42, 26 July 2014

Problem

Given that $2^{2004}$ is a $604$-digit number whose first digit is $1$, how many elements of the set $S = \{2^0,2^1,2^2,\ldots ,2^{2003}\}$ have a first digit of $4$?

$\mathrm{(A)}\ 194 \qquad \mathrm{(B)}\ 195 \qquad \mathrm{(C)}\ 196 \qquad \mathrm{(D)}\ 197 \qquad \mathrm{(E)}\ 198$

Solution

Given $n$ digits, there must be exactly one power of $2$ with $n$ digits such that the first digit is $1$. Thus $S$ contains $603$ elements with a first digit of $1$. For each number in the form of $2^k$ such that its first digit is $1$, then $2^{k+1}$ must either have a first digit of $2$ or $3$, and $2^{k+2}$ must have a first digit of $4,5,6,7$. Thus there are also $603$ numbers with first digit $\{2,3\}$ and $603$ numbers with first digit $\{4,5,6,7\}$. By using complementary counting, there are $2004 - 3 \times 603 = 195$ elements of $S$ with a first digit of $\{8,9\}$. Now, $2^k$ has a first of $\{8,9\}$ if and only if the first digit of $2^{k-1}$ is $4$, so there are $\boxed{195} \Rightarrow \mathrm{(B)}$ elements of $S$ with a first digit of $4$.

Alternate Solution

We can make the following chart for the possible loops of leading digits: \[1 \rightarrow 2 \rightarrow 4 \rightarrow 8 \rightarrow 1\] \[1 \rightarrow 2 \rightarrow 4 \rightarrow 9 \rightarrow 1\] \[1 \rightarrow 2 \rightarrow 5 \rightarrow 1\] \[1 \rightarrow 3 \rightarrow 6 \rightarrow 1\] \[1 \rightarrow 3 \rightarrow 7 \rightarrow 1\]

Let there be $x$ of the sequences of $4$ numbers, and let there be $y$ of the sequences of $3$ numbers. We note that a $4$ appears only in the loops of $4$, and also we are given that $2^{2004}$ has $604$ digits. \[3x+4y=2004\] \[x+y=603\] Solving gives $x = 408$ and $y = 195$, thus the answer is $(B)$.

See also

2004 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last problem
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=2004_AMC_12B_Problems/Problem_25&oldid=62690"