2023年12月10日发(作者：小学生数学试卷可爱名称)

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2004 AMC 10B

Problem 1

Each row of the Misty Moon Amphitheater has 33

seats. Rows 12 through 22 are reserved for a youth

club. How many seats are reserved for this club?

There are

seats.

Problem 2

How many two-digit positive integers have at

least one 7 as a digit?

Ten numbers (Nine numbers () have as the tens digit.

) have it as the ones digit.

rows of seats, giving

Number is in both sets.

Thus the result is

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

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.

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48 free throws. How many free throws did she make

at the first practice?

At the fourth practice she made

the third one it was

throws, at

, then we get

throws for the second practice, and finally

throws at the first one.

Problem 4

A standard six-sided die is rolled, and P is the

product of the five numbers that are visible. What

is the largest number that is certain to divide

P?

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, can not 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 .

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. __________________________________________________

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

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?

If

If

Case

or , the expression evaluates to

.

.

.

, the expression evaluates to

remains.

In that case, we want to maximize where .

Trying out the six possibilities we get that the best one is

, where

Problem 6

Which of the following numbers is a perfect square?

Using the fact that

▪

▪

▪

▪

▪

.

, we can write:

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Clearly is a square, and as , , and are primes,

none of the other four are squares.

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

Solution 2

Each time Isabelle exchanges U.S. dollars, she gets Canadian

dollars and Canadian dollars extra. Isabelle received a total of

Canadian dollars extra, therefore she exchanged U.S. dollars

times. Thus

Problem 8

Minneapolis-St. Paul International Airport is 8 miles southwest of

downtown St. Paul and 10 miles southeast of downtown Minneapolis.

Which of the following is closest to the number of miles between

downtown St. Paul and downtown Minneapolis?

The directions \"southwest\" and \"southeast\" are orthogonal. Thus the

described situation is a right triangle with legs 8 miles and 10 miles

.

, and the sum of digits of is

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long. The hypotenuse length is

is .

, and thus the answer

Without a calculator one can note that

Problem 9

.

A square has sides of length 10, and a circle centered at one of its

vertices has radius 10. What is the area of the union of the regions

enclosed by the square and the circle?

The area of the circle is

Exactly

, the area of the square is .

of the circle lies inside the square. Thus the total area is

.

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?

The sum of the first odd numbers is

have

Problem 11

Two eight-sided dice each have faces numbered 1 through 8. 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?

.

. As in our case , we

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Solution 1

We have , hence if at least one of the numbers is ,

the sum is larger. There such possibilities.

We have .

, hence all other cases

the sum is

it is smaller.

For we already have

are good.

Out of the possible cases, we found that in

greater than or equal to the product, hence in

Therefore the answer is

Solution 2

Let the two rolls be

From the restriction:

Since

either

and

, and .

.

are non-negative integers between and

,

if and only if or

with

and

, or

.

, ordered pairs with

. So, there are

.

and

,

There are ordered pairs

and ordered pair with

ordered pairs such that

,

if and only if

and

or equivalently

. . This gives ordered pair

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So, there are a total of

.

Since there are a total of

ordered pairs with

ordered pairs with

ordered pairs

.

, there are

Thus, the desired probability is

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?

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The area of the large circle is

hence the shaded area is

, the area of the small one is

.

we have

.

,

From the for the right triangle

and thus the shaded area is

Problem 13

, hence

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?

All numbers in this solution will be in hundreds of a millimeter.

The thinnest coin is the dime, with thickness

has height .

. A stack of dimes

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 the stack will be too low and if

we are left with cases and .

If the possible stack heights are

remaining ones exceeding .

Therefore there are coins in the stack.

it will be too high. Thus

, with the

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 ).

Problem 14

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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?

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 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

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

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of all marbles in the bag. __________________________________________________

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

nickels dimes. As the total count is

and dimes.

Problem 16

, this means that there are

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?

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 . Therefore the radius we seek is

.

WARNING. Note that the answer does not correspond to any of the

five options. Most probably there is a typo in option D.

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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

.

In five years, Jack\'s age will be

.

We are given that

, then Bill\'s current age is

and Bill\'s age will be

. Thus .

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For we get

integer, and for

. For and the value is not an

it is more than . Thus the only solution is

. , and the difference in ages is

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 , hence it is a

multiple of . Thus Bill\'s current age modulo must be .

Thus Bill\'s age is in the set .

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 .

Checking each of them, we see that only

solution

Problem 18

In the right triangle , we have , , and .

Points , , and are located on , , and , respectively, so

that , , and . What is the ratio of the area of

to that of ?

.

works, and gives the

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First of all, note that

.

Draw the height from onto as in the picture below:

, and therefore

Now consider the area of . Clearly the triangles and

, are similar, as they have all angles equal. Their ratio is

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hence . Now the area

=

of can be computed as

.

Similarly we can find that as well.

Hence , and the answer is

.

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 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: and

. This is easily proved by induction. It follows that

.

Solution 2

Note that the recurrence

.

can be rewritten as

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Hence we get that

and also

From the values given in the problem statement we see that

.

From

From

Following

we get that

we get that

this pattern,

.

Problem 20

In points

intersect at

?

and lie on and , respectively. If

and

and

.

.

we get

so that , what is

Solution (Triangle Areas)

We use the square bracket notation to denote area.

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Without loss of generality, we can assume

, and . We have

.

. Then

,

so we need to find the area of quadrilateral

Draw the line segment

. Let

and

triangles

get ,

.

to form the two triangles and

, and

, we obtain

and , we obtain

. By considering triangles

, and by considering

. Solving, we

is , so the area of quadrilateral

Therefore

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 . Then, to balance and on , has a mass of .

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

mass of to .

.

), we assign a mass of to and a

Now, to balance and on , we must give a mass of

Finally, the ratio of to is given by the ratio of the mass of to

. the mass of , which is

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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

way we will have

the picture below:

, and

, , and . This

. The situation is shown in

The point is the intersection of the lines

the first line have the form

have the form

.

The ratio can now be computed simply by observing the

coordinates of , , and :

and . The points on

, the points on the second line

, hence . Solving for we get

Problem 21

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Let ; ; and ; ;

the union of the first

numbers are in ?

be two arithmetic progressions. The set is

terms of each sequence. How many distinct

The two sets of terms are

.

Now . We can compute

and

. We will now find

.

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

.

Thus the good values of are

.

Therefore

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?

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, and

, the second one gives

, and their count is

, and thus . __________________________________________________

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 .

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

, where is the area of the triangle and

and ,

well-known identity

its perimeter. In our case,

thus

be at

The distance

. As the inscribed circle touches both legs, its center must

.

of these two points

.

is then

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?

Label the six sides of the cube by numbers to as on a classic dice.

Then the \"four vertical faces\" can be:

.

Let be the set of colorings where are all of the same color,

similarly let and be the sets of good colorings for the other two

sets of faces.

, , or

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There are possible colorings, and there are good

colorings. Thus the result is

.

Using the we

. We need to compute

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

.

Problem 24

In we have , , and

circumscribed circle of the triangle so that

the value of ?

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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. Point

bisects

is on the

. What is __________________________________________________

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 and to get the picture below:

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Now obviously the triangles

side .

and are equilateral with

Take a look at the bottom circle. The angle is , hence the

sector is of the circle. The same is true for the sector of

the bottom circle, and sectors and of the top circle.

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.

Hence the area of the new shaded region is

, and the area of the

original shared region is

.

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