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2004 AMC 10B Problems/Problem 1
Each row of the Misty Moon Amphitheater has seats. Rows through
reserved for a youth club. How many seats are reserved for this club?
are
Solution
There are
rows of
seats, giving
seats.
2004 AMC 10B Problems/Problem 2
How many two-digit positive integers have at least one
as a digit?
Solution 1
Ten numbers
have as the tens digit. Nine numbers
as the ones digit. Number
is in both sets.
Thus the result is
have it
.
Solution 2
We use complementary counting. The complement of having at least one
having no s as a digit.
We have digits to choose from for the first digit and
a total of
two-digit numbers.
But since we cannot have
choose from.
Thus there are
digits for the second. This gives
as a digit, we have first digits and second digits to
two-digit numbers without a
(The total number of two-digit numbers)
without a )
as a digit is
as a digit.
(The number of two-digit numbers
.
2004 AMC 10B Problems/Problem 3
At each basketball practice last week, Jenny made twice as many free throws as she
made at the previous practice. At her fifth practice she made free throws. How many
free throws did she make at the first practice?
Solution
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At the fourth practice she made
we get
throws, at the third one it was
throws for the second practice, and finally
, then
throws at
the first one.
2004 AMC 10B Problems/Problem 4
A standard six-sided die is rolled, and is the product of the five numbers that are
visible. What is the largest number that is certain to divide ?
Solution 1
The product of all six numbers is
. The products of numbers that can be visible
are
,
, ...,
. The answer to this problem is their greatest common divisor - which is
, where is the least common multiple of
.
Clearly
and the answer is
.
Solution 2
Clearly,
cannot have a prime factor other than , and .
We can not guarantee that the product will be divisible by , as the number can end
on the bottom.
We can guarantee that the product will be divisible by (one of and will always be
visible), but not by .
Finally, there are three even numbers, hence two of them are always visible and thus
the product is divisible by . This is the most we can guarantee, as when the is on the
bottom side, the two visible even numbers are and , and their product is not divisible
by .
Hence
.
2004 AMC 10B Problems/Problem 5
In the expression
, the values of , , , and are , , , and , although not
necessarily in that order. What is the maximum possible value of the result?
Solution
If
If
or
, the expression evaluates to
, the expression evaluates to
.
.
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Case
remains. In that case, we want to maximize
Trying out the six possibilities we get that the best one is
where
where
.
,
.
2004 AMC 10B Problems/Problem 6
Which of the following numbers is a perfect square?
Solution
Using the fact that
, we can write:
We see that
is a square, and because
, and
are primes, none of the
other four choices are squares.
2004 AMC 10B Problems/Problem 7
On a trip from the United States to Canada, Isabella took U.S. dollars. At the border
she exchanged them all, receiving Canadian dollars for every U.S. dollars. After
spending Canadian dollars, she had Canadian dollars left. What is the sum of the
digits of ?
Solution 1
Isabella had
Canadian dollars. Setting up an equation we get
which solves to
, and the sum of digits of
is
,
.
Solution 2
Each time Isabella exchanges U.S. dollars, she gets Canadian dollars and Canadian
dollars extra. Isabella received a total of Canadian dollars extra, therefore she
exchanged
U.S. dollars
times. Thus
2004 AMC 10B Problems/Problem 8
.
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Minneapolis-St. Paul International Airport is miles southwest of downtown St. Paul
and miles southeast of downtown Minneapolis. Which of the following is closest to
the number of miles between downtown St. Paul and downtown Minneapolis?
Solution
The directions "southwest" and "southeast" are orthogonal. Thus the described
situation is a right triangle with legs miles and miles long. The hypotenuse length
is
, and thus the answer is
.
Without a calculator one can note that
.
2004 AMC 10B Problems/Problem 9
A square has sides of length , and a circle centered at one of its vertices has radius
What is the area of the union of the regions enclosed by the square and the circle?
Solution
The area of the circle is
Exactly
is
; the area of the square is
.
of the circle lies inside the square. Thus the total area
.
2004 AMC 10B Problems/Problem 10
A grocer makes a display of cans in which the top row has one can and each lower row
has two more cans than the row above it. If the display contains
cans, how many
rows does it contain?
Solution
.
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The sum of the first
odd numbers is
. As in our case
, we have
.
2004 AMC 10B Problems/Problem 11
Two eight-sided dice each have faces numbered through . When the dice are rolled,
each face has an equal probability of appearing on the top. What is the probability that
the product of the two top numbers is greater than their sum?
Solution 1
We have
, hence if at least one of the numbers is , the sum is larger.
There such possibilities.
We have
For
.
we already have
, hence all other cases are good.
Out of the
possible cases, we find that in
the sum is greater than or
equal to the product, hence in
cases the sum is smaller, satisfying the
condition. Therefore the answer is
.
Solution 2
Let the two rolls be
, and .
From the restriction:
Since
either
and
are non-negative integers between and ,
,
, or
if and only if
or
There are ordered pairs
with
pair with
and
. So, there are
that
.
if and only if
This gives ordered pair
.
, ordered pairs with
ordered pairs
and
.
or equivalently
, and ordered
such
and
.
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So, there are a total of
ordered pairs
with
Since there are a total of
pairs
with
ordered pairs
.
, there are
Thus, the desired probability is
.
ordered
.
2004 AMC 10B Problems/Problem 12
An annulus is the region between two concentric circles. The concentric circles in the
figure have radii and , with
. Let
be a radius of the larger circle, let
be
tangent to the smaller circle at , and let
be the radius of the larger circle that
contains . Let
,
, and
. What is the area of the annulus?
Solution
The area of the large circle is
is
.
, the area of the small one is
From the Pythagorean Theorem for the right triangle
hence
and thus the shaded area is
, hence the shaded area
we have
.
2004 AMC 10B Problems/Problem 13
In the United States, coins have the following thicknesses: penny,
mm;
nickel,
mm; dime,
mm; quarter,
mm. If a stack of these coins is
exactly mm high, how many coins are in the stack?
Solution
,
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All numbers in this solution will be in hundredths of a millimeter.
The thinnest coin is the dime, with thickness
. A stack of
dimes has height
.
The other three coin types have thicknesses
,
, and
. By replacing
some of the dimes in our stack by other, thicker coins, we can clearly create exactly all
heights in the set
.
If we take an odd , then all the possible heights will be odd, and thus none of them will
be
. Hence is even.
If
cases
the stack will be too low and if
and
.
it will be too high. Thus we are left with
If
the possible stack heights are
exceeding
.
Therefore there are
, with the remaining ones
coins in the stack.
Using the above observation we can easily construct such a stack. A stack of dimes
would have height
, thus we need to add
. This can be done for example
by replacing five dimes by nickels (for
), and one dime by a penny (for
).
Note
We can easily add up
get
and
to get
. We multiply that by to
. Since this works and it requires 8 coins, the answer is clearly
.
2004 AMC 10B Problems/Problem 14
A bag initially contains red marbles and blue marbles only, with more blue than red.
Red marbles are added to the bag until only
of the marbles in the bag are blue. Then
yellow marbles are added to the bag until only of the marbles in the bag are blue.
Finally, the number of blue marbles in the bag is doubled. What fraction of the marbles
now in the bag are blue?
Solution
We can ignore most of the problem statement. The only important information is that
immediately before the last step blue marbles formed of the marbles in the bag. This
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means that there were blue and
other marbles, for some . When we double the
number of blue marbles, there will be
blue and
other marbles, hence blue
marbles now form
of all marbles in the bag.
2004 AMC 10B Problems/Problem 15
Patty has coins consisting of nickels and dimes. If her nickels were dimes and her
dimes were nickels, she would have cents more. How much are her coins worth?
Solution 1
She has nickels and
dimes. Their total cost
is
cents. If the dimes were nickels and vice versa,
she would have
cents. This value should
be cents more than the previous one. We get
, which solves
to
. Her coins are worth
.
Solution 2
Changing a nickel into a dime increases the sum by cents, and changing a dime into a
nickel decreases it by the same amount. As the sum increased by cents, there
are
more nickels than dimes. As the total count is , this means that there
are
nickels and dimes, which is equal to
.
2004 AMC 10B Problems/Problem 16
Three circles of radius are externally tangent to each other and internally tangent to a
larger circle. What is the radius of the large circle?
Solution
The situation in shown in the picture below. The radius we seek is
Clearly
. The point is clearly the center of the equilateral triangle
thus
is
of the altitude of this triangle. We get that
radius we seek is
.
.
,
. Therefore the
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2004 AMC 10B Problems/Problem 17
The two digits in Jack's age are the same as the digits in Bill's age, but in reverse order.
In five years Jack will be twice as old as Bill will be then. What is the difference in their
current ages?
Solution 1
If Jack's current age is
, then Bill's current age is
In five years, Jack's age will be
We are given that
For
for
ages is
.
and Bill's age will be
. Thus
.
.
we get
. For
and
the value
it is more than . Thus the only solution is
is not an integer, and
, and the difference in
.
Solution 2
Age difference does not change in time. Thus in five years Bill's age will be equal to
their age difference.
The age difference is
current age modulo must be .
Thus Bill's age is in the set
, hence it is a multiple of . Thus Bill's
.
As Jack is older, we only need to consider the cases where the tens digit of Bill's age is
smaller than the ones digit. This leaves us with the options
.
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Checking each of them, we see that only
solution
works, and gives the
.
2004 AMC 10B Problems/Problem 18
In the right triangle
, we have
,
, and
and are located on
,
, and
, respectively, so that
and
. What is the ratio of the area of
to that of
. Points
,
,
,
,
?
Solution 1
Let
triangles,
Because of
. Because
is divided into four
.
triangle
area,
.
and
, so
, so
.
.
Solution 2
First of all, note that
Draw the height from
, and therefore
onto
as in the picture below:
.
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Now consider the area of
. Clearly the triangles
they have all angles equal. Their ratio is
area
of
can be computed
as
and
are similar, as
, hence
=
. Now the
.
Similarly we can find that
as well.
Hence
, and the answer is
.
2004 AMC 10B Problems/Problem 19
In the sequence
,
,
,
, each term after the third is found by subtracting
the previous term from the sum of the two terms that precede that term. For example,
the fourth term is
. What is the
term in this sequence?
Solution
Solution 1
We already know that
,
,
, and
. Let's compute the
next few terms to get the idea how the sequence behaves. We
get
,
,
, and so on.
We can now discover the following pattern:
is easily proved by induction. It follows that
and
.
Solution 2
Note that the recurrence
as
can be rewritten
.
. This
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Hence we get that
also
and
From the values given in the problem statement we see that
From
we get that
.
From
we get that
.
.
Following this pattern, we get
.
2004 AMC 10B Problems/Problem 20
In
at
points
so that
and
lie on
and
and
, respectively. If
, what is
?
and
intersect
Solution (Triangle Areas)
We use the square bracket notation
to denote area.
Without loss of generality, we can assume
and
. We have
quadrilateral
.
. Then
,
, so we need to find the area of
Draw the line segment
to form the two triangles
and
.
Let
, and
. By considering triangles
and
obtain
, and by considering triangles
and
, we
obtain
. Solving, we get
,
, so the area of
quadrilateral
is
.
Therefore
, we
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Solution (Mass points)
The presence of only ratios in the problem essentially cries out for mass points.
As per the problem, we assign a mass of to point , and a mass of to
balance and on , has a mass of .
. Then, to
Now, were we to assign a mass of to and a mass of to , we'd have . Scaling this
down by
(to get , which puts and in terms of the masses of and ), we
assign a mass of
to
Now, to balance
and
Finally, the ratio of
and a mass of
on
to
which is
to
, we must give
.
a mass of
is given by the ratio of the mass of
.
to the mass of
,
.
Solution (Coordinates)
Affine transformations preserve ratios of distances, and for any pair of triangles there is
an affine transformation that maps the first one onto the second one. This is why the
answer is the same for any
, and we just need to compute it for any single
triangle.
We can choose the points
have
, and
,
, and
. This way we will
. The situation is shown in the picture below:
The point is the intersection of the lines
and
. The points on the first line have
the form
, the points on the second line have the form
. Solving
for we get
, hence
.
The ratio
, and :
can now be computed simply by observing the
coordinates of
,
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2004 AMC 10B Problems/Problem 21
Let ; ;
and ; ;
be two arithmetic progressions. The set is the union of the
first
terms of each sequence. How many distinct numbers are in ?
Solution
Solution 1
The two sets of terms are
and
.
Now
. We can compute
will now find
.
. We
Consider the numbers in . We want to find out how many of them lie in . In other
words, we need to find out the number of valid values of for which
.
The fact "
" can be rewritten as "
The first condition gives
, the second one gives
Thus the good values of are
Therefore
, and
.
, and their count is
, and thus
".
.
.
Solution 2
Shift down the first sequence by and the second by so that the two sequences
become
and
. The first becomes multiples of and the
second becomes multiples of . Their intersection is the multiples of up to
.
There are
multiples of
. There are
distinct
numbers in .
2004 AMC 10B Problems/Problem 22
A triangle with sides of 5, 12, and 13 has both an inscribed and a circumscribed circle.
What is the distance between the centers of those circles?
Solution
This is obviously a right triangle. Pick a coordinate system so that the right angle is
at
and the other two vertices are at
and
.
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As this is a right triangle, the center of the circumcircle is in the middle of the
hypotenuse, at
.
The radius of the inscribed circle can be computed using the well-known
identity
, where is the area of the triangle and its perimeter. In our
case,
and
, thus
. As the inscribed circle
touches both legs, its center must be at
.
The distance of these two points is then
.
2004 AMC 10B Problems/Problem 23
Each face of a cube is painted either red or blue, each with probability 1/2. The color of
each face is determined independently. What is the probability that the painted cube
can be placed on a horizontal surface so that the four vertical faces are all the same
color?
Solution
Label the six sides of the cube by numbers to as on a classic dice. Then the "four
vertical faces" can be:
,
, or
.
Let
let
be the set of colorings where
are all of the same color, similarly
and be the sets of good colorings for the other two sets of faces.
There are
result is
possible colorings, and there are
. We need to compute
good colorings. Thus the
.
Using the Principle of Inclusion-Exclusion we can write
Clearly
, as we have two possibilities for the common color of the
four vertical faces, and two possibilities for each of the horizontal faces.
What is
? The faces
must have the same color, and at the same time
faces
must have the same color. It turns out
that
the set containing just the two cubes where
all six faces have the same color.
Therefore
, and the result is
.
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Solution 2
Suppose we break the situation into cases that contain four vertical faces of the same
color:
I. Two opposite sides of same color There are 3 ways to choose the two sides, and then
two colors possible, so
II. One face different from all the others There are 6 ways to choose this face, and 2
colors, so
III. All faces same There are 2 colors, so two ways for all faces to be the same.
Adding them up, we have a total of 20 ways to have four vertical faces the same color.
The are
ways to color the cube, so the answer is
2004 AMC 10B Problems/Problem 24
In triangle
we have
circle of the triangle so that
,
,
bisects angle
. Point is on the circumscribed
. What is the value of
?
Solution
Set
's length as .
's length must also be
since
and
of equal length. Using Ptolemy's Theorem,
intercept arcs
. The ratio is
Solution 2
Let
Furthermore,
Then
. Observe that
, so
because they subtend the same arc.
is similar to
by AAA similarity.
. By angle bisector theorem,
so
gives
. Plugging this into the similarity proportion
gives:
.
which
2004 AMC 10B Problems/Problem 25
A circle of radius is internally tangent to two circles of radius at points and ,
where
is a diameter of the smaller circle. What is the area of the region, shaded in
the picture, that is outside the smaller circle and inside each of the two larger circles?
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Solution
The area of the small circle is . We can add it to the shaded region, compute the area of
the new region, and then subtract the area of the small circle from the result.
Let and be the intersections of the two large circles. Connect them to
get the picture below:
Now obviously the triangles
and
are equilateral with side .
and
to
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Take a look at the bottom circle. The angle
the circle. The same is true for the sector
sectors
and
of the top circle.
is , hence the sector
of the bottom circle, and
is
of
If we now sum the areas of these four sectors, we will almost get the area of the new
shaded region - except that each of the two equilateral triangles will be counted twice.
These triangles have a base of and a height of .
Hence the area of the new shaded region is
and the area of the original shared region is
,
.