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

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Revision as of 09:57, 10 February 2008

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 a 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 either $\{2,3\}$ or $\{4,5,6,7\}$. By the complement principle, 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 $4$.

See also

2004 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last problem
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All AMC 12 Problems and Solutions
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