美国数学竞赛2020年真题AMC10 A(真题加详解)

2023年12月31日发(作者：深圳中学数学试卷)

2020 AMC 10A Solution

Problem1

What value of satisfies

Solution

Adding to both

sides, .

Problem2

The numbers and have an average (arithmetic mean) of .

What is the average of and ?

Solution

The arithmetic mean of the numbers and is equal

to

we get

. Solving for

. Dividing by to find the average of the two

,

numbers and gives .

Problem3

Assuming , , and , what is the value in simplest form of

the following expression?

Solution

Note that is times .

Likewise, is times and

Therefore, the product of the given fraction

equals .

is times .

Problem4

A driver travels for hours at

gets

miles per hour, during which her car

per mile, and her only miles per gallon of gasoline. She is paid

expense is gasoline at

per hour, after this expense?

per gallon. What is her net rate of pay, in dollars

Solution

Since the driver travels 60 miles per hour and each hour she uses 2 gallons of

gasoline, she spends $4 per hour on gas. If she gets $0.50 per mile, then she

gets $30 per hour of driving. Subtracting the gas cost, her net rate of pay per

hour is .

Problem5

What is the sum of all real numbers for which

Solution 1

Split the equation into two cases, where the value inside the absolute value is

positive and nonpositive.

Case 1:

The equation yields

to

is and .

Case 2:

Similarly, taking the nonpositive case for the value inside the absolute value

notation yields

gives

Summing all the values results in .

. Factoring and simplifying

, so the only value for this case is .

, which is equal

. Therefore, the two values for the positive case

Solution 2

We have the

equations and .

Notice that the second is a perfect square with a double root at , and the

first has real roots. By Vieta\'s, the sum of the roots of the first equation

is .

Problem6

How many -digit positive integers (that is, integers between

inclusive) having only even digits are divisible by

and ,

Solution

The ones digit, for all numbers divisible by 5, must be either or . However,

from the restriction in the problem, it must be even, giving us exactly one choice

() for this digit. For the middle two digits, we may choose any even integer

from , meaning that we have total options. For the first digit, we follow

similar intuition but realize that it cannot be , hence giving us 4 possibilities.

Therefore, using the multiplication rule, we

get . ~ciceronii

Problem7

The integers from to inclusive, can be arranged to form a -by- square in which the sum of the numbers in each row, the sum of the numbers

in each column, and the sum of the numbers along each of the main diagonals

are all the same. What is the value of this common sum?

Solution

Without loss of generality, consider the five rows in the square. Each row must

have the same sum of numbers, meaning that the sum of all the numbers in the

square divided by is the total value per row. The sum of the integers

is , and the

common sum is

.

Solution 2

Take the sum of the middle 5 values of the set (they will turn out to be the mean

of each row). We get

~Baolan

as our answer.

Problem8

What is the value of

Solution 1

Split the even numbers and the odd numbers apart. If we group every 2 even

numbers together and add them, we get a total of .

Summing the odd numbers is equivalent to summing the first 100 odd numbers,

which is equal to

of .

. Adding these two, we obtain the answer

Solution 2 (bashy)

We can break this entire sum down into integer bits, in which the sum is ,

where is the first integer in this bit. We can find that the first sum of every

sequence is , which we plug in for the bits in the entire sequence

is

term of every sequence equation we got

above , and so the sum of every bit is ,

, so then we can plug it into the first

and we only found the value of , the sum of the sequence

is .-middletonkids

Solution 3

Another solution involves adding everything and subtracting out what is not

needed. The first step involves

solving

. To do this, we can simply multiply and and divide by to get

us . The next step involves subtracting out the numbers with minus

signs. We actually have to do this twice, because we need to take out the

numbers we weren’t supposed to add and then subtract them from the problem.

Then, we can see that from to , incrementing by , there are numbers

that we have to subtract. To do this we can do times divided by , and

then we can multiply by , because we are counting by fours, not ones. Our

answer will be , but remember, we have to do this twice. Once we do that,

we will get

answer is

—

. Finally, we just have to do

.

, and our

Solution 4

In this solution, we group every 4 terms. Our groups should

be: ,

, ...

. We add them together to get this

expression:

as

get . ~Baolan

. This can be rewritten

. We add this to

,

Solution 5

We can split up this long sum into groups of four integers. Finding the first few

sums, we have that

and

, ,

. Notice that this is an increasing arithmetic

sequence, with a common difference of . We can find the sum of the arithmetic

sequence by finding the average of the first and last terms, and then multiplying

by the number of terms in the sequence. The first term is

or , the last term is

are or

, or , and there

,

terms. So, we have that the sum of the sequence

is

, or . ~Arctic_Bunny

Solution 3

Taking the average of the first and last terms, and , we have that the

mean of the set is . There are 5 values in each row, column or diagonal, so the

value of the common sum is

KINGLOGIC

, or . ~Arctic_Bunny, edited by

Problem9

A single bench section at a school event can hold either adults or children.

When bench sections are connected end to end, an equal number of adults

and children seated together will occupy all the bench space. What is the least

possible positive integer value of

Solution

The least common multiple of and is . Therefore, there must

be adults and children. The total number of benches

is

.

Solution 2

This is similar to Solution 1, with the same basic idea, but we don\'t need to

calculate the LCM. Since both and are prime, their LCM must be their

product. So the answer would be . ~Baolan

Problem10

Seven cubes, whose volumes are , , , , , , and cubic

units, are stacked vertically to form a tower in which the volumes of the cubes

decrease from bottom to top. Except for the bottom cube, the bottom face of each

cube lies completely on top of the cube below it. What is the total surface area of

the tower (including the bottom) in square units?

Solution 1

The volume of each cube follows the pattern of

between and .

ascending, for is

We see that the total surface area can be comprised of three parts: the sides of

the cubes, the tops of the cubes, and the bottom of the cube

(which is just ). The sides areas can be measured as the

sum , giving us . Structurally, if we examine the tower from the

square of area . Therefore, we

.

top, we see that it really just forms a

can say that the total surface area is

Alternatively, for the area of the tops, we could have found the

sum

~ciceronii

, giving us as well.

Solution 2

It can quickly be seen that the side lengths of the cubes are the integers from 1 to

7, inclusive.

First, we will calculate the total surface area of the cubes, ignoring overlap. This

value

is

. Then, we need to subtract out the overlapped parts of the cubes. Between each

consecutive pair of cubes, one of the smaller cube\'s faces is completely covered,

along with an equal area of one of the larger cube\'s faces. The total area of the

overlapped parts of the cubes is thus equal to

the overlapped surface area from the total surface area, we

get . ~emerald_block

. Subtracting

Solution 3 (a bit more tedious than others)

It can be seen that the side lengths of the cubes using cube roots are all integers

from to , inclusive.

Only the cubes with side length and have faces in the surface area and

the rest have . Also, since the

cubes are stacked, we have to find the difference between

each

.

We then come up with

this:

.

We then add all of this and get .

and side length as ranges from to

Problem 11

What is the median of the following list of numbers

Solution 1

We can see that

the

is less than 2020. Therefore, there are

. Also, there are

is equal to

of

numbers after

. Since

numbers that are under

, it, with the other squares, will and equal to

shift our median\'s placement up

set is , and

is

~aryam

.

. We can find that the median of the whole

gives us . Our answer

Solution 2

As we are trying to find the median of a -term set, we must find the

average of the th and st terms.

Since is slightly greater than , we know that

the perfect squares through are less than , and the rest are

, there greater. Thus, from the number to the number

are

than and

terms. Since

less than

is less

, we will only need to consider the

perfect square terms going down from the th term, , after going

down terms. Since the th and st terms are

only and terms away from the th term, we can simply

subtract from and from to get the two terms, which

are and . Averaging the two, we

get ~emerald_block

Solution 3

We want to know the

We know that

So numbers

So the sum of and

the th number.

Also, notice that

Then the

the

th term will be

th term will be .

th term and the

are in between to

will result in

.

is

th term to get the median.

, which means that

.

, and similarly

, which is larger than

Solving for the median of the two numbers, we get

Problem12

Triangle

Medians

and

is isoceles with

and

.

are perpendicular to each other,

. What is the area of

Solution 1

Since quadrilateral has perpendicular diagonals, its area can be

found as half of the product of the length of the diagonals. Also note

that has the area of triangle by similarity,

so Thus,

Solution 2 (Trapezoid)

We know that , and since the ratios of its sides

are , the ratio of of their areas is .

If

area of

is the area of

.

and

, then trapezoid is the

Let\'s call the intersection of

Then

of triangles

have base .

. Let

, and

.

are heights . Since

and , respectively. Both of these triangles

Area of

Area of

Adding these two gives us the area of trapezoid

is .

, which

This is of the triangle, so the area of the triangle

is ~quacker88, diagram by programjames1

Solution 3 (Medians)

Draw median .

Since we know that all medians of a triangle intersect at the incenter, we know

that passes through point . We also know that medians of a triangle

. Knowing this, we can see divide each other into segments of ratio

that

to , and

, and since the two segments sum

are and , respectively.

Finally knowing that the medians divide the triangle into sections of equal area,

finding the area of is enough. .

The area of

us

~quacker88

. Multiplying this by gives

Solution 4 (Triangles)

We know

that

As

that

ratio of

So

, , so

, we can see

and

.

with a side

.

, and the area of

.

,

With that, we can see that

trapezoid is 72.

As said in solution 1,

-QuadraticFunctions, solution 1 by ???

.

Solution 5 (Only Pythagorean Theorem)

Let be the height. Since medians divide each other into a ratio, and

the medians have length 12, we

have and . From right

triangle

so

From right

triangle

which implies

,

. By

,

. Since is a median, .

symmetry .

Applying the Pythagorean Theorem to right

triangle gives

,

so

of is

. Then the area

Solution 6 (Drawing)

(NOT recommended) Transfer the given diagram, which happens to be to scale,

onto a piece of a graph paper. Counting the boxes should give a reliable result

since the answer choices are relatively far apart. -Lingjun

Solution 7

Given a triangle with perpendicular medians with lengths and , the area will

be .

Solution 8 (Fastest)

Connect the line segment and it\'s easy to see

quadrilateral has an area of the product of its diagonals divided

by which is . Now, solving for triangle could be an option, but the

drawing shows the area of will be less than the quadrilateral meaning

the the area of is less than but greater than , leaving only

one possible answer choice, .

Problem 13

A frog sitting at the point begins a sequence of jumps, where each jump

is parallel to one of the coordinate axes and has length , and the direction of

each jump (up, down, right, or left) is chosen independently at random. The

sequence ends when the frog reaches a side of the square with

vertices and . What is the probability that the

sequence of jumps ends on a vertical side of the square

Solution

Drawing out the square, it\'s easy to see that if the frog goes to the left, it will

immediately hit a vertical end of the square. Therefore, the probability of this

happening is . If the frog goes to the right, it will be in the center of

the square at , and by symmetry (since the frog is equidistant from all

sides of the square), the chance it will hit a vertical side of a square is . The

probability of this happening is .

If the frog goes either up or down, it will hit a line of symmetry along the corner it

is closest to and furthest to, and again, is equidistant relating to the two closer

sides and also equidistant relating the two further sides. The probability for it to

hit a vertical wall is . Because there\'s a chance of the frog going up and

down, the total probability for this case is and summing up all the

cases,

Solution 2

Let\'s say we have our four by four grid and we work this out by casework. A is

where the frog is, while B and C are possible locations for his second jump, while

O is everything else. If we land on a C, we have reached the vertical side.

However, if we land on a B, we can see that there is an equal chance of reaching

the horizontal or vertical side, since we are symmetrically between them. So we

have the probability of landing on a C is 1/4, while B is 3/4. Since C means that

we have \"succeeded\", while B means that we have a half chance, we

compute

.

We get , or

-yeskay

Solution 3

If the frog is on one of the 2 diagonals, the chance of landing on vertical or

horizontal each becomes . Since it starts on , there is a chance (up,

down, or right) it will reach a diagonal on the first jump and chance (left) it will

reach the vertical side. The probablity of landing on a vertical

is - Lingjun.

Solution 4 (Complete States)

Let denote the probability of the frog\'s sequence of jumps ends with it

. Note that

.

, and .

by hitting a vertical edge when it is at

reflective symmetry over the line

Similarly,

Now we create equations for the probabilities at each of these points/states by

considering the probability of going either up, down, left, or right from that

point:

We have a system of equations

in variables, so we can solve for each of these probabilities. Plugging the

second equation into the fourth equation

gives

Plugging in the third equation into this

gives

Next, plugging

in the second and third equation into the first equation

yields

Now plugging in (*) into this, we

get

Problem14

Real numbers and satisfy and . What is the

value of

Solution

Continuing to

combine

From the givens, it can be concluded that .

Also,

that

This means

. Substituting this information

into , we

have . ~PCChess

Solution 2

As above, we need to calculate

the roots of

so

Thus

and

and .

. Note that are

where and as in the previous solution. Thus the

answer is

.

Solution 3

Note that Now, we

only need to find the values of

Recall that

that

and

and

We are able to solve the

second equation, and doing so gets us

into the first equation, we get

Plugging this

In order to find the value of

we can add them together. This gets

we find a common denominator so that

us

that

get

Recalling

and solving this equation, we

Plugging this into the first equation, we

get

Solving the original equation, we

get

~emerald_block

Solution 4 (Bashing)

This is basically bashing using Vieta\'s formulas to find and (which I highly do

not recommend, I only wrote this solution for fun).

We use Vieta\'s to find a quadratic relating and . We set and to be the

roots of the quadratic

and

roots

(because ,

). We can solve the quadratic to get the

and . and are \"interchangeable\", meaning that it

doesn\'t matter which solution or is, because it\'ll return the same result when

plugged in. So we plug in

get as our answer.

for and and

~Baolan

Solution 5 (Bashing Part 2)

This usually wouldn\'t work for most problems like this, but we\'re lucky that we can

quickly expand and factor this expression in this question.

We first change the original expression to

because

to

. This is equal

,

. We can factor and

reduce to

. Now our expression is just

factor

would be

to get

.

. We

. So the answer

Solution 6 (Complete Binomial Theorem)

We first simplify the expression to Then, we can solve

for and given the system of equations in the problem.

Since we can substitute for . Thus, this becomes the

equation

obtain

we obtain

Multiplying both sides by , we

or

. We also easily find that

By the quadratic formula

given , equals the conjugate of . Thus, plugging our values

in for and , our expression equalsBy the binomial theorem, we observe that every second terms of the

expansions and will cancel out (since a positive plus a negative of the

same absolute value equals zero). We also observe that the other terms not

canceling out are doubled when summing the expansions of . Thus,

our expression equalswhich

equalswhich equals .

Problem15

A positive integer divisor of is chosen at random. The probability that the

divisor chosen is a perfect square can be expressed as

relatively prime positive integers. What is ?

, where and are

Solution

The prime factorization of is . This yields a total

of divisors of In order to produce a perfect square

divisor, there must be an even exponent for each number in the prime

factorization. Note that and can not be in the prime factorization of a

perfect square because there is only one of each in Thus, there

are perfect squares. (For , you can have , , , , , or 0 s,

etc.) The probability that the divisor chosen is a perfect square

is

Problem16

A point is chosen at random within the square in the coordinate plane whose

vertices are and . The

is probability that the point is within units of a lattice point is . (A point

a lattice point if and are both integers.) What is to the nearest tenth

Solution 1

Diagram

Diagram by MathandSki Using Asymptote

Note: The diagram represents each unit square of the

given square.

Solution

We consider an individual one-by-one block.

If we draw a quarter of a circle from each corner (where the lattice points are

located), each with radius , the area covered by the circles should be .

Because of this, and the fact that there are four circles, we write

Solving for , we obtain

and from here, we simplify and see

that

, where with , we get ,

~Crypthes

To be more rigorous, note that since if then

clearly the probability is greater than . This would make sure the above solution

works, as if

quartercircles.

there is overlap with the

Solution 2

As in the previous solution, we obtain the equation ,

which simplifies to . Since is slightly more than , is

slightly less than

than

. We notice that

, so is roughly

is slightly more

~emerald_block

Solution 3 (Estimating)

As above, we find that we need to estimate

Note that we can approximate

.

and

so

.

.

And so our answer is

Problem 17

Defineintegers are there such that ?

How many

Solution 1

Notice that is a product of many integers. We either need one factor to be

0 or an odd number of negative factors.

Case 1: There are 100 integers for which

Case 2: For there to be an odd number of negative factors, must be between

an odd number squared and an even number squared. This means that there

are

are

total possible values of . Simplifying, there

possible numbers.

total possible values of . ~PCChess

Summing, there are

Solution 2

Notice that

between

is nonpositive when is

and , and , and (inclusive), which

means that the amount of values

equals

.

This reduces

to

~Zeric

Solution 3 (end behavior)

We know that

coefficient. That

is,

Since the degree of

.

is even, its end behaviors match. And since the

as goes in

is a -degree function with a positive leading

leading coefficient is positive, we know that both ends approach

either direction.

So the first time is going to be negative is when it intersects the -axis at

, which is the an -intercept and it\'s going to dip below. This happens at

smallest intercept.

However, when it hits the next intercept, it\'s going to go back up again into

positive territory, we know this happens at . And when it hits , it\'s going to

dip back into negative territory. Clearly, this is going to continue to snake around

the intercepts until .

To get the amount of integers below and/or on the -axis, we simply need to

count the integers. For example, the amount of integers in between

the interval we got earlier, we subtract and add

one. integers, so there are four integers in this interval

that produce a negative result.

Doing this with all of the other intervals, we have

.

Proceed with Solution 2. ~quacker88

Problem18

Let be an ordered quadruple of not necessarily distinct integers,

For how many such quadruples is it true

is one such quadruple,

each one of them in the set

that

because

is odd? (For example,

is odd.)

Solution

Solution 1 (Parity)

In order for to be odd, consider parity. We must have (even)-(odd) or (odd)-(even). There are ways to pick numbers

to obtain an even product. There are ways to obtain an odd product.

Therefore, the total amount of ways to make odd

is

-Midnight

.

Solution 2 (Basically Solution 1 but more in depth)

Consider parity. We need exactly one term to be odd, one term to be even.

Because of symmetry, we can set to be odd and to be even, then multiply

by If is odd, both and must be odd, therefore there

are possibilities for Consider Let us say that is even.

Then there are possibilities for However, can be odd, in which

case we have more possibilities for Thus there are ways for

us to choose and ways for us to choose Therefore, also considering

symmetry, we have total values of

Solution 3 (Complementary Counting)

There are 4 ways to choose any number independently and 2 ways to choose

any odd number independently. To get an even products, we

count:

which is

be counted like so:

,

. The number of ways to get an odd product can

, which is , or . So, for one

product to be odd the other to be even:

matters). ~ Anonymous and Arctic_Bunny

(order

Solution 4 (Solution 3 but more in depth)

We use complementary counting: If the difference is even, then we can subtract

those cases. There are a total of cases.

For an even difference, we have (even)-(even) or (odd-odd).

From Solution 3:

\"There are 4 ways to choose any number independently and 2 ways to choose

any odd number independently. even products:(number)*(number)-(odd)*(odd): . odd products: (odd)*(odd): .\"

With this, we easily calculate .

Problem19

As shown in the figure below, a regular dodecahedron (the polyhedron consisting

of congruent regular pentagonal faces) floats in space with two horizontal

faces. Note that there is a ring of five slanted faces adjacent to the top face, and

a ring of five slanted faces adjacent to the bottom face. How many ways are there

to move from the top face to the bottom face via a sequence of adjacent faces so

that each face is visited at most once and moves are not permitted from the

bottom ring to the top ring?

Diagram

Solution 1

Since we start at the top face and end at the bottom face without moving from the

lower ring to the upper ring or revisiting a face, our journey must consist of the

top face, a series of faces in the upper ring, a series of faces in the lower ring,

and the bottom face, in that order.

We have choices for which face we visit first on the top ring. From there, we

have choices for how far around the top ring we go before moving

down: or faces around clockwise, or faces around

counterclockwise, or immediately going down to the lower ring without visiting

any other faces in the upper ring.

We then have choices for which lower ring face to visit first (since every upper-ring face is adjacent to exactly lower-ring faces) and then once again choices

for how to travel around the lower ring. We then proceed to the bottom face,

completing the trip.

Multiplying together all the numbers of choices we have, we

get .

Solution 2

Swap the faces as vertices and the vertices as faces. Then, this problem is the

same as 2016 AIME I #3which had an answer

of .

Problem20

Quadrilateral satisfies

and

ls and intersect at point

and

Diagona What is the area of

quadrilateral

Solution 1 (Just Drop An Altitude)

It\'s crucial to draw a good diagram for this one.

Since

need to find

and , we get . Now we

to get the area of the whole quadrilateral. Drop an altitude

from to and call the point of intersection . Let .

Since , then . By dropping this altitude, we can also

see two similar triangles, and .

Since is , and , we get that .

Now, if we redraw another diagram just of , we get

that

dividing by the GCF, we get

to

Since

So

,

. Now expanding, simplifying, and

. This factors

. . Since lengths cannot be negative,

.

(I\'m very sorry if you\'re a visual learner but now you have a diagram by ciceronii)

~ Solution by Ultraman

~ Diagram by ciceronii

Solution 2 (Pro Guessing Strats)

We know that the big triangle has area 300. Use the answer choices which would

mean that the area of the little triangle is a multiple of 10. Thus the product of the

legs is a multiple of 20. Guess that the legs are equal to

and because the hypotenuse is 20 we get

numbers, we get that when and ,

and

. Testing small

is indeed a square. The

.

,

area of the triangle is thus 60, so the answer is

~tigershark22 ~(edited by HappyHuman)

Solution 3 (coordinates)

Let the points

be , ,

lies on line

,

, we know that

,

,and

.

,

,

respectively. Since

Furthermore, since

so

solutions

lies on the circle with diameter

. Solving for and with these equations, we get the

and . We immediately discard

the

that

solution as should be negative. Thus, we conclude

.

Solution 4 (Trigonometry)

Let and Using Law of Sines

on

on

we get

yields

and LoS

Divide the

two to

get Now,

and

solve the quadratic, taking the positive solution (C is acute) to

get So

if then and By Pythagorean

Theorem,

is

and the answer

(This solution is incomplete, can someone complete it please-Lingjun) ok Latex

edited by kc5170

We could use the famous m-n rule in trigonometry in triangle ABC with Point E

[Unable to write it anybody write the expression] We will find that BD

is angle bisector of triangle ABC(because we will get tan (x)=1) Therefore by

converse of angle bisector theorem AB:BC = 1:3. By using phythagorean

theorem we have values of AB and AC. = 120. Adding area of ABC and

ACD Answer••360

Problem21

There exists a unique strictly increasing sequence of nonnegative

integers such

thatWhat is

Solution 1

First, substitute with . Then, the given equation

becomes

consider only

equals

that equals

. This

. Note

. Now

, since the sum of a geometric

sequence is . Thus, we can see that forms the sum of 17

different powers of 2. Applying the same method to each

of , , ... , , we can see that each of the pairs

. But forms the sum of 17 different powers of 2. This gives us

we must count also the

is

~seanyoon777

term. Thus, Our answer

.

Solution 2

(This is similar to solution 1) Let

be rewritten

as

. Then, . The LHS can

. Plugging

have

back in for , we

. When expanded, this will have

answer is .

terms. Therefore, our

Solution 3 (Intuitive)

Multiply both sides by

get

Notice that

extra term of

two

, since there is a on the LHS. However, now we have an

on the right from . To cancel it, we let

, so we let

. The

to

\'s now combine into a term of . And so on,

.

.

. So we

until we get to . Now everything we don\'t want telescopes into

We already have that term since we let

Everything from now on will automatically telescope to

let be .

As you can see, we will have to add

automatically telescope for the next

\'s at a time, then \"wait\" for the sum to

. We numbers, etc, until we get to

only need to add \'s between odd multiples of

largest even multiple of below is

total of

and even multiples. The

, so we will have to add a

at the

.

\'s. However, we must not forget we let

beginning, so our answer is

Solution 4

Note that the expression is equal to something slightly lower than

answer choices

for terms is

and

.

. Clearly,

make no sense because the lowest sum

just makes no sense. and are 1

, and apart, but because the expression is odd, it will have to contain

because

~Lcz

is bigger, the answer is .

Solution 5

In order to shorten expressions, will represent consecutive s when

expressing numbers.

Think of the problem in binary. We have

Note that

and

Since

this means that

so

Expressing each of the pairs of the form

or

This means that each pair has terms of the form

Since there are of these pairs, there are a total of

Accounting for the

of

.

terms.

in binary, we have

term, which was not in the pair, we have a total

terms.

Problem22

For how many positive

integers

by ? (Recall that

isnot divisible

is the greatest integer less than or equal to .)

Solution 1 (Casework)

Expression:Solution:

Let

Since , for any integer , the difference between

the largest and smallest terms before the function is applied is less than or

equal to , and thus the terms must have a range of or less after the function is

applied.

This means that for every integer ,

if is an integer and

,

, then the three terms in the expression

above must be

if is an integer because , then will be an integer and will

be greater than

be ,

; thus the three terms in the expression must

if

be

is an integer, then the three terms in the expression above must

,

if

be

is an integer, then the three terms in the expression above must

, and

if none of

expression above must be .

are integral, then the three terms in the

The last statement is true because in order for the terms to be different, there

must be some integer in the interval or the

interval . However, this means that multiplying the integer

by should produce a new integer between and or and ,

exclusive, but because no such integers exist, the terms cannot be different, and

thus, must be equal.

Note that does not work; to prove this, we just have to

substitute for in the expression. This gives

us

which is divisible by 3.

Now, we test the five cases listed above (where

Case 1: divides and

, so the

)

As mentioned above, the three terms in the expression are

sum is , which is divisible by . Therefore, the first case does not work

(0 cases).

Case 2: divides and

, which As mentioned above, in this case the terms must be

means the sum is

this is 1 case that works.

Case 3: divides

Because divides

number of factors of

=

, so the expression is not divisible by . Therefore,

, the number of possibilities for is the same as the

.

is . . So, the total number of factors of

However, we have to subtract , because the case

mentioned previously. This leaves 7 cases.

Case 4: divides

Because divides

number of factors of

=

does not work, as

, the number of possibilities for is the same as the

.

is . . So, the total number of factors of

Again, we have to subtract , so this leaves cases. We have

also overcounted the factor , as it has been counted as a factor of and

as a separate case (Case 2). , so there are actually 14 valid

cases.

Case 5: divides none of

. The Similar to Case 1, the value of the terms of the expression are

sum is , which is divisible by 3, so this case does not work (0 cases).

Now that we have counted all of the cases, we add them.

, so the answer is

~dragonchomper, additional edits by emerald_block

.

Solution 2 (Solution 1 but simpler)

Note that this solution does not count a majority of cases that are important to

consider in similar problems, though they are not needed for this problem, and

therefore it may not work with other, similar problems.

Notice that you only need to count the number of factors of 1000 and 999,

excluding 1. 1000 has 16 factors, and 999 has 8. Adding them gives you 24, but

you need to subtract 2 since 1 does not work.

Therefore, the answer is 24 - 2 = .

-happykeeper, additional edits by dragonchomper

Solution 3

NOTE: For this problem, whenever I say

factors of the number except for .

Now, quickly observe that if divides ,

, I will be referring to all the

then and will also round down to , giving us a sum

of

if

, which does not work for the question. However,

divides , we see

that and . This gives us a

sum of

logic, we can deduce that

we need the factors of

because the

when

only that satisfies that is

have:

, which is clearly not divisible by . Using the same

too works (for our problem). Thus,

and and we don\'t have to eliminate any

. But we have to be careful! See that

, then our problem doesn\'t get fulfilled. The

. So, we

;

. Adding them up gives a total of workable \'s.

Problem23

Let be the triangle in the coordinate plane with

and Consider the following five isometries vertices

(rigid transformations) of the plane: rotations

of and counterclockwise around the origin, reflection across

the -axis, and reflection across the -axis. How many of the sequences of

three of these transformations (not necessarily distinct) will return to its

original position? (For example, a rotation, followed by a reflection across

the -axis, followed by a reflection across the -axis will return to its original

position, but a rotation, followed by a reflection across the -axis, followed

by another reflection across the -axis will not return to its original position.)

Solution

First, any combination of motions we can make must reflect an even number

of times. This is because every time we reflect , it changes orientation.

Once has been flipped once, no combination of rotations will put it back in

place because it is the mirror image; however, flipping it again changes it back to

the original orientation. Since we are only allowed transformations and an even

number of them must be reflections, we either reflect times or times.

Case 1: 0 reflections on T

In this case, we must use rotations to return

that our set of rotations,

of

in

one

rotations which ensures that

except for

to its original position. Notice

, contains every multiple

. We can start with any two rotations

and there must be exactly

such that we can use the three

.

yields a full rotation. For example, That way, the composition of rotations

if

then

so and the rotations

,

,

yields a full rotation.

. This happens The only case in which this fails is when would have to equal

when

namely,

is already a full rotation,

or .

However, we can simply subtract these three cases from the total.

Selecting from yields choices, and

with that fail, we are left with combinations for case 1.

Case 2: 2 reflections on T

In this case, we first eliminate the possibility of having two of the same reflection.

Since two reflections across the x-axis maps back to itself, inserting a rotation

before, between, or after these two reflections would change \'s final location,

meaning that any combination involving two reflections across the x-axis would

not map back to itself. The same applies to two reflections across the y-axis.

Therefore, we must use one reflection about the x-axis, one reflection about the

y-axis, and one rotation. Since a reflection about the x-axis changes the sign of

the y component, a reflection about the y-axis changes the sign of the x

component, and a rotation changes both signs, these three transformation

composed (in any order) will suffice. It is therefore only a question of arranging

the three, giving us combinations for case 2.

Combining both cases we get

Solution 2(Rewording solution 1)

As in the previous solution, note that we must have either 0 or 2 reflections

because of orientation since reflection changes orientation that is impossible to

fix by rotation. We also know we can\'t have the same reflection twice, since that

would give a net of no change and would require an identity rotation.

Suppose there are no reflections. Denote as 1, as 2, and as

3, just for simplification purposes. We want a combination of 3 of these that will

sum to either 4 or 8(0 and 12 is impossible since the minimum is 3 and the max is

9). 4 can be achieved with any permutation of

achieved with any permutation of

in ways.

and 8 can be

. This case can be done

Suppose there are two reflections. As noted already, they must be different, and

as a result will take the triangle to the opposite side of the origin if we don\'t do

any rotation. We have 1 rotation left that we can do though, and the only one that

will return to the original position is 2, which is AKA reflection across

origin. Therefore, since all 3 transformations are distinct. The three

transformations can be applied anywhere since they are commutative(think

quadrants). This gives ways.

Problem24

Let be the least positive integer greater than

which for

What is the sum of the digits of ?

Solution 1

We know that

write

get

write

these two modular congruences,

, so we can

. Simplifying, we

. Similarly, we can

, or . Solving

which we know is

the only solution by CRT (Chinese Remainder Theorem). Now, since the problem

is asking for the least positive integer greater than , we find the least

solution is . However, we are have not considered cases

where or

so we

try

o again we add

that

conditions, so our answer is

.

to . It turns out

does indeed satisfy the original

.

s.

Solution 2 (bashing)

We are given

that

This tells us that

that

of

and

is divisible by but not . It also tells us

.

is divisible by 60 but not 120. Starting, we find the least value

which is divisible by which satisfies the conditions for , which

is , making . We then now keep on adding until we get a

number which satisfies the second equation. This number turns out to be ,

whose digits add up to

-Midnight

.

Solution 3 (bashing but worse)

Assume that has 4 digits. Then

digits of the number (not to get confused with

problem,

So we know that

that

and

(last digit of ). That means

and . We

such

, where , , , represent

). As given the

.

can bash this after this. We just want to find all pairs of numbers

that is a multiple of 7 that is greater than a multiple of . Our equation

for

for

would be

would be

and our equation

, where is any integer. We plug

this value in until we get a value of that makes satisfy the

original problem statement (remember, ). After bashing for

hopefully a couple minutes, we find that works.

So

~ Baolan

which means that the sum of its digits is .

Solution 4

The conditions of the problem reduce to the

following.

where

that

that ,

where and

. From these equations, we see

. Solving this diophantine equation gives us

form. Since, is greater than ,

we can do some bounding and get that and . Now we start

the bash by plugging in numbers that satisfy these conditions. We

get , . So the answer is .

Solution 5

You can first find that n must be congruent

to and . The we can find

that and , where x and y are integers.

Then we can find that y must be odd, since if it was even the gcd will be 120, not

60. Also, the unit digit of n has to be 7, since the unit digit of 60y is always 0 and

the unit digit of 57 is 7. Therefore, you can find that x must end in 1 to satisfy n

having a unit digit of 7. Also, you can find that x must not be a multiple of three or

else the gcd will be 63. Therefore, you can test values for x and you can find that

x=91 satisfies all these ore, n is 1917 and

=.-happykeeper

Solution 6 (Reverse Euclidean Algorithm)

We are given

that and By

applying the Euclidean algorithm, but in reverse, we

have

and

We now know that

by

integer

be

Since

then

We know that

must be divisible by

Therefore,

and so it is divisible

for some

will

).

or else the first condition won\'t hold ( will be ) and or else the second condition won\'t hold ( gives us too small of an answer,

so the answer

is

Problem25

Jason rolls three fair standard six-sided dice. Then he looks at the rolls and

chooses a subset of the dice (possibly empty, possibly all three dice) to reroll.

After rerolling, he wins if and only if the sum of the numbers face up on the three

dice is exactly Jason always plays to optimize his chances of winning. What is

the probability that he chooses to reroll exactly two of the dice?

Solution 1

Consider the probability that rolling two dice gives a sum of , where

There are

namely

pairs that satisfy this,

, out

.

of possible pairs. The probability is .

Therefore, if one die has a value of and Jason rerolls the other two dice, then

the probability of winning is .

In order to maximize the probability of winning, must be minimized. This means

that if Jason rerolls two dice, he must choose the two dice with the maximum

values.

Thus, we can let

call

If

, , and

be the values of the three dice, which we will

respectively. Consider the case when .

, then we do not need to reroll any dice. Otherwise, if we

in the hope that we get the value that makes reroll one die, we can roll dice

the sum of the three dice . This happens with probability . If we reroll two

dice, we will roll

above.

and , and the probability of winning is , as stated

However,

if .

, so rolling one die is always better than rolling two dice