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Difference between revisions of "2004 AMC 12B Problems/Problem 25"

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== Problem ==
 
== Problem ==
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Given that <math>2^{2004}</math> is a <math>604</math>-[[digit]] number whose first digit is <math>1</math>, how many [[element]]s of the [[set]] <math>S = \{2^0,2^1,2^2,\ldots ,2^{2003}\}</math> have a first digit of <math>4</math>?  
 
Given that <math>2^{2004}</math> is a <math>604</math>-[[digit]] number whose first digit is <math>1</math>, how many [[element]]s of the [[set]] <math>S = \{2^0,2^1,2^2,\ldots ,2^{2003}\}</math> have a first digit of <math>4</math>?  
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<math>\mathrm{(A)}\ 194
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\qquad\mathrm{(B)}\ 195
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<math>\mathrm{(A)}\ 194 \qquad \mathrm{(B)}\ 195 \qquad \mathrm{(C)}\ 196 \qquad \mathrm{(D)}\ 197 \qquad \mathrm{(E)}\ 198</math>
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\qquad\mathrm{(C)}\ 196
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== Solution 1==
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\qquad\mathrm{(D)}\ 197
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\qquad\mathrm{(E)}\ 198</math>
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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 digit 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>.
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== Solution ==
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Given <math>n</math> digits, there must be a 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 either <math>\{2,3\}</math> or <math>\{4,5,6,7\}</math>. By the [[complement principle]], 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>.  
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== Solution 2 ==
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We can make the following chart for the possible loops of leading digits:
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<cmath>1 \rightarrow 2 \rightarrow 4 \rightarrow 8 \rightarrow 1</cmath>
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<cmath>1 \rightarrow 2 \rightarrow 4 \rightarrow 9 \rightarrow 1</cmath>
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<cmath>1 \rightarrow 2 \rightarrow 5 \rightarrow 1</cmath>
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<cmath>1 \rightarrow 3 \rightarrow 6 \rightarrow 1</cmath>
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<cmath>1 \rightarrow 3 \rightarrow 7 \rightarrow 1</cmath>
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 +
 
 +
Thus each loop from <math>1 \rightarrow 1</math> can either have <math>3</math> or <math>4</math> numbers. Let there be <math>x</math> of the sequences of <math>3</math> numbers, and let there be <math>y</math> of the sequences of <math>4</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.
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<cmath>3x+4y=2004</cmath>
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<cmath>x+y=603</cmath>
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Solving gives <math> x = 408</math> and <math>y = 195</math>, thus the answer is <math>(B)</math>.
  
 
== See also ==
 
== See also ==
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[[Category:Intermediate Number Theory Problems]]
 
[[Category:Intermediate Number Theory Problems]]
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{{MAA Notice}}

Latest revision as of 19:12, 15 October 2016

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 1

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 digit 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 $4$.

Solution 2

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


Thus each loop from $1 \rightarrow 1$ can either have $3$ or $4$ numbers. Let there be $x$ of the sequences of $3$ numbers, and let there be $y$ of the sequences of $4$ 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)
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