2024年4月11日发(作者：青岛初二统考期中数学试卷)

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USA AMC 10 2000

1

In the year , the United States will host the International

Mathematical Olympiad. Let , , and be distinct positive integers

such that the product . What\'s the largest possible

value of the sum ?

Solution

The sum is the highest if two factors are the lowest.

So,

2

and .

Solution

.

3

Each day, Jenny ate of the jellybeans that were in her jar at the

beginning of the day. At the end of the second day, remained. How

many jellybeans were in the jar originally?

Solution

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4

Chandra pays an online service provider a fixed monthly fee plus an

hourly charge for connect time. Her December bill was , but in

January her bill was because she used twice as much connect

time as in December. What is the fixxed monthly fee?

Solution

Let be the fixed fee, and be the amount she pays for the minutes

she used in the first month.

We want the fixed fee, which is

5

Points and are the midpoints of sides

moves along a line that is parallel to side

quantities listed below change?

(a) the length of the segment

(b) the perimeter of

(c) the area of

and of . As

, how many of the four

(d) the area of trapezoid

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Solution

(a) Clearly does not change, and , so doesn\'t

change either.

(b) Obviously, the perimeter changes.

(c) The area clearly doesn\'t change, as both the base

corresponding height remain the same.

and its

(d) The bases and do not change, and neither does the height,

so the trapezoid remains the same.

Only quantity changes, so the correct answer is .

6

The Fibonacci Sequence starts with two 1s and

each term afterwards is the sum of its predecessors. Which one of the

ten digits is the last to appear in thet units position of a number in the

Fibonacci Sequence?

Solution

The pattern of the units digits are

In order of appearance:

.

is the last.

7

In rectangle , , is on , and

. What is the perimeter of ?

and trisect

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Solution

.

Since

Thus,

.

Adding,

8

At Olympic High School, of the freshmen and of the sophomores

took the AMC-10. Given that the number of freshmen and sophomore

contestants was the same, which of the following must be true?

There are five times as many sophomores as freshmen.

There are twice as many sophomores as freshmen.

There are as many freshmen as sophomores.

There are twice as many freshmen as sophomores.

There are five times as many freshmen as sophomores.

Solution

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is trisected,

.

.

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Let be the number of freshman and be the number of sophomores.

There are twice as many freshmen as sophomores.

9

If

, where , then

Solution

, so

.

.

.

10

The sides of a triangle with positive area have lengths , , and .

The sides of a second triangle with positive area have lengths , ,

and . What is the smallest positive number that is not a possible

value of ?

Solution

From the triangle inequality,

positive number not possible is

and

, which is .

. The smallest

11

Two different prime numbers between and are chosen. When their

sum is subtracted from their product, which of the following numbers

could be obtained?

Solution

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Two prime numbers between and are both odd.

Thus, we can discard the even choices.

Both and

of four.

are even, so one more than is a multiple

is the only possible choice.

satisfy this,

12

Figures , , , and consist of , , , and nonoverlapping unit

squares, respectively. If the pattern were continued, how many

nonoverlapping unit squares would there be in figure 100?

.

Solution

Solution 1

We have a recursion:

.

I.E. we add increasing multiples of each time we go up a figure.

So, to go from Figure 0 to 100, we add

.

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Solution 2

We can divide up figure to get the sum of the sum of the first

odd numbers and the sum of the first odd numbers. If you do not see

this, here is the example for :

The sum of the first odd numbers is

unit squares. We plug in

choice

13

There are 5 yellow pegs, 4 red pegs, 3 green pegs, 2 blue pegs, and 1

orange peg to be placed on a triangular peg board. In how many ways

can the pegs be placed so that no (horizontal) row or (vertical) column

contains two pegs of the same color?

, so for figure , there are

to get , which is

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Solution

In each column there must be one yellow peg. In particular, in the

rightmost column, there is only one peg spot, therefore a yellow peg

must go there.

In the second column from the right, there are two spaces for pegs.

One of them is in the same row as the corner peg, so there is only one

remaining choice left for the yellow peg in this column.

By similar logic, we can fill in the yellow pegs as shown:

After this we can proceed to fill in the whole pegboard, so there is only

arrangement of the pegs. The answer is

14

Mrs. Walter gave an exam in a mathematics class of five students.

She entered the scores in random order into a spreadsheet, which

recalculated the class average after each score was entered. Mrs.

Walter noticed that after each score was entered, the average was

always an integer. The scores (listed in ascending order) were , ,

, , and . What was the last score Mrs. Walter entered?

Solution

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The sum of the first scores must be even, so we must choose evens

or the odds to be the first two scores.

Let us look at the numbers in mod .

If we choose the two odds, the next number must be a multiple of ,

of which there is none.

Similarly, if we choose or , the next number must be a

multiple of , of which there is none.

So we choose first.

remains. The next number must be 1 in mod 3, of which only

The sum of the first three scores is

.

. This is equivalent to in mod

is the only Thus, we need to choose one number that is in mod .

one that works.

Thus, is the last score entered.

15

Two non-zero real numbers, and , satisfy

following is a possible value of

?

. Which of the

Solution

Substituting , we get

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16

The diagram shows lattice points, each one unit from its nearest

neighbors. Segment meets segment at . Find the length of

segment .

Solution

Solution 1

Let be the line containing and and let be the line containing

and . If we set the bottom left point at

, , and .

. The -intercept is

, then ,

The line is given by the equation

, so . We are given two points on , hence we can

to be , so is the line

. The slope in this case is

gives us

,

compute the slope,

Similarly, is given by

so

line

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. Plugging in the point

.

, so is the


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