Difference between revisions of "2004 AMC 12B Problems/Problem 25"
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== Solution == | == Solution == | ||
− | 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 | + | 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 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>. |
== See also == | == See also == |
Revision as of 14:34, 18 September 2011
Problem
Given that is a -digit number whose first digit is , how many elements of the set have a first digit of ?
Solution
Given digits, there must be a power of with digits such that the first digit is . Thus contains elements with a first digit of . For each number in the form of such that its first digit is , then must either have a first digit of or , and must have a first digit of . Thus there are also numbers with first digit either or . By using complementary counting, there are elements of with a first digit of . Now, has a first of if and only if the first digit of is , so there are elements of with a first digit of .
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 |