Difference between revisions of "2004 AMC 12B Problems/Problem 25"
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− | 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> | + | 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>3x+4y=2004</cmath> | ||
<cmath>x+y=603</cmath> | <cmath>x+y=603</cmath> |
Revision as of 23:58, 17 March 2015
Contents
[hide]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 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 of
if and only if the first digit of
is
, so there are
elements of
with a first digit of
.
Alternate Solution
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.