Difference between revisions of "2004 AMC 12B Problems/Problem 25"
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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> | <math>\mathrm{(A)}\ 194 \qquad \mathrm{(B)}\ 195 \qquad \mathrm{(C)}\ 196 \qquad \mathrm{(D)}\ 197 \qquad \mathrm{(E)}\ 198</math> | ||
− | == Solution == | + | == Solution 1== |
− | Given <math>n</math> digits, there must be | + | 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>. |
+ | |||
+ | == Solution 2 == | ||
+ | |||
+ | 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> | ||
+ | |||
+ | |||
+ | 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. | ||
+ | <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 == | ||
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[[Category:Intermediate Number Theory Problems]] | [[Category:Intermediate Number Theory Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 19:12, 15 October 2016
Contents
Problem
Given that is a -digit number whose first digit is , how many elements of the set have a first digit of ?
Solution 1
Given digits, there must be exactly one 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 and numbers with first digit . By using complementary counting, there are elements of with a first digit of . Now, has a first digit of if and only if the first digit of is , so there are elements of with a first digit of .
Solution 2
We can make the following chart for the possible loops of leading digits:
Thus each loop from can either have or numbers. Let there be of the sequences of numbers, and let there be of the sequences of numbers. We note that a appears only in the loops of , and also we are given that has digits.
Solving gives and , thus the answer is .
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 |
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