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

Revision as of 18:50, 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: $1 \rightarrow 2 \rightarrow 4 \rightarrow 8 \rightarrow 1$ $1 \rightarrow 2 \rightarrow 4 \rightarrow 9 \rightarrow 1$ $1 \rightarrow 2 \rightarrow 5 \rightarrow 1$ $1 \rightarrow 3 \rightarrow 6 \rightarrow 1$ $1 \rightarrow 3 \rightarrow 7 \rightarrow 1$

Thus each loop from $1 \rightarrow 1$ can either have $3$ or $4$ numbers. 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 $(B)$ .

@@ Line 22: / Line 22: @@
 <cmath>x+y=603</cmath>
 Solving gives <math> x = 408</math> and <math>y = 195</math>, thus the answer is <math>(B)</math>.
---[[User:Az phx brandon jiang|Az phx brandon jiang]] 19:49, 26 July 2014 (EDT)
 == See also ==

2004 AMC 12B (Problems • Answer Key • Resources)
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Difference between revisions of "2004 AMC 12B Problems/Problem 25"

Revision as of 18:50, 26 July 2014

Contents

Problem

Solution

Alternate Solution

See also