Difference between revisions of "2004 AMC 12B Problems/Problem 25"
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== Solution 1== | == Solution 1== | ||
− | 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 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>. | + | 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: | We can make the following chart for the possible loops of leading digits: |
Latest revision as of 20:12, 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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