2020 AMC10A真题及答案
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参考答案见文末（仅供参考）
Problem 1
What value of satisfies
Problem 2
The numbers and have an average (arithmetic mean) of . What is the average of and ?
Problem 3
Assuming , , and , what is the value in simplest form of the following expression?
Problem 4
A driver travels for hours at miles per hour, during which her car gets miles per gallon of gasoline. She is paid per mile, and her only expense is gasoline at per gallon. What is her net rate of pay, in dollars per hour, after this expense?
Problem 5
What is the sum of all real numbers for which
Problem 6
How many digit positive integers (that is, integers between and , inclusive) having only even digits are divisible by
Problem 7
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?
Problem 8
What is the value of
Problem 9
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
Problem 10
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?
Problem 11
What is the median of the following list of numbers
Problem 12
Triangle is isosceles with . Medians and are perpendicular to each other, and . What is the area of
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
Problem 14
Real numbers and satisfy and . What is the value of
Problem 15
A positive integer divisor of is chosen at random. The probability that the divisor chosen is a perfect square can be expressed as , where and are relatively prime positive integers. What is ?
Problem 16
A point is chosen at random within the square in the coordinate plane whose vertices are and . The probability that the point is within units of a lattice point is . (A point is a lattice point if and are both integers.) What is to the nearest tenth
Problem 17
DefineHow many integers are there such that ?
Problem 18
Let be an ordered quadruple of not necessarily distinct integers, each one of them in the set For how many such quadruples is it true that is odd? (For example, is one such quadruple, because is odd.)
Problem 19
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?
Problem 20
Quadrilateral satisfies and Diagonals and intersect at point and What is the area of quadrilateral
Problem 21
There exists a unique strictly increasing sequence of nonnegative integers such thatWhat is
Problem 22
For how many positive integers isnot divisible by ? (Recall that is the greatest integer less than or equal to .)
Problem 23
Let be the triangle in the coordinate plane with vertices and Consider the following five isometries (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.)
Problem 24
Let be the least positive integer greater than for whichWhat is the sum of the digits of ?
Problem 25
Jason rolls three fair standard sixsided 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?
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 Adding to both sides, .
 The arithmetic mean of the numbers and is equal to . Solving for , we get . Dividing by to find the average of the two numbers and gives .
 Note that is times . Likewise, is times and is times . Therefore, the product of the given fraction equals .
 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 .

Solution 1
Split the equation into two cases, where the value inside the absolute value is positive and nonpositive.
Case 1:
The equation yields , which is equal to . Therefore, the two values for the positive case is and .
Case 2:
Similarly, taking the nonpositive case for the value inside the absolute value notation yields . Factoring and simplifying gives , so the only value for this case is .
Summing all the values results in .
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 . .
 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 .
 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 as our answer.
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 , or .

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 . Adding these two, we obtain the answer of .
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 , so then we can plug it into the first term of every sequence equation we got above , and so the sum of every bit is , and we only found the value of , the sum of the sequence is .
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 . Finally, we just have to do , and our answer is .
Solution 4
In this solution, we group every 4 terms. Our groups should be: , , , ... . We add them together to get this expression: . This can be rewritten as . We add this to get .
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 , or , and there are or terms. So, we have that the sum of the sequence is , or .

Solution 1
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 .

Solution 1
The volume of each cube follows the pattern of ascending, for is between and .
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 top, we see that it really just forms a square of area . Therefore, we can say that the total surface area is . Alternatively, for the area of the tops, we could have found the sum , 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 . Subtracting the overlapped surface area from the total surface area, we get .
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 and side length as ranges from to
.
We then come up with this: .
We then add all of this and get .

Solution 1
We can see that is less than 2020. Therefore, there are of the numbers after . Also, there are numbers that are under and equal to . Since is equal to , it, with the other squares, will shift our median's placement up . We can find that the median of the whole set is , and gives us . Our answer is .
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 greater. Thus, from the number to the number , there are terms. Since is less than and less than , 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
Solution 3
We want to know the th term and the th term to get the median.
We know that
So numbers are in between to .
So the sum of and will result in , which means that is the th number.
Also, notice that , which is larger than .
Then the th term will be , and similarly the th term will be .
Solving for the median of the two numbers, we get 
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 is the area of , then trapezoid is the area of .
Let's call the intersection of and . Let . Then . Since , and are heights of triangles and , respectively. Both of these triangles have base .
Area of
Area of
Adding these two gives us the area of trapezoid , which is .
This is of the triangle, so the area of the triangle is
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 divide each other into segments of ratio . Knowing this, we can see that , and since the two segments sum to , and 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 . Multiplying this by gives us
Solution 4 (Triangles)
We know that , , so .
As , we can see that and with a side ratio of .
So , .
With that, we can see that , and the area of trapezoid is 72.
As said in solution 1, .
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 . Since is a median, . From right triangle ,which implies . By symmetry .
Applying the Pythagorean Theorem to right triangle gives , so . Then the area of is
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, .

Solution 1
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
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
Solution 4 (Complete States)
Let denote the probability of the frog's sequence of jumps ends with it hitting a vertical edge when it is at . Note that by reflective symmetry over the line . Similarly, , and . 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 givesPlugging in the third equation into this givesNext, plugging in the second and third equation into the first equation yieldsNow plugging in (*) into this, we get

Solution 1
Continuing to combineFrom the givens, it can be concluded that . Also,This means that . Substituting this information into , we have . ~PCChess
Solution 2
As above, we need to calculate . Note that are the roots of and so and . Thus where and as in the previous solution. Thus the answer is .
Solution 3
Note that Now, we only need to find the values of and
Recall that and that We are able to solve the second equation, and doing so gets us Plugging this into the first equation, we get
In order to find the value of we find a common denominator so that we can add them together. This gets us Recalling that and solving this equation, we get Plugging this into the first equation, we get
Solving the original equation, we get
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 (because , and ). We can solve the quadratic to get the roots 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 for and and get asour answer.
~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 . This is equal to . We can factor and reduce to . Now our expression is just . We factor to get . So the answer would be .
Solution 6 (Complete Binomial Theorem)
We first simplify the expression toThen, we can solve for and given the system of equations in the problem. Since we can substitute for . Thus, this becomes the equationMultiplying both sides by , we obtain orBy the quadratic formula we obtain . We also easily find that 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 .
 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

Solution 1
Diagram
Diagram by MathandSki Using Asymptote
Note: The diagram represents each unit square of the given square.
Solution
We consider an individual onebyone 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 , where with , we get , and from here, we simplify and see that ~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 there is overlap with the quartercircles.
Solution 2
As in the previous solution, we obtain the equation , which simplifies to . Since is slightly more than , is slightly less than . We notice that is slightly more than , so is roughly
Solution 3 (Estimating)
As above, we find that we need to estimate .
Note that we can approximate and so .
And so our answer is .

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 total possible values of . Simplifying, there are possible numbers.
Summing, there are total possible values of . ~PCChess
Solution 2
Notice that is nonpositive when is between and , and , and (inclusive), which means that the amount of values equals .
This reduces to
~Zeric
Solution 3 (end behavior)
We know that is a degree function with a positive leading coefficient. That is, .
Since the degree of is even, its end behaviors match. And since the leading coefficient is positive, we know that both ends approach as goes in either direction.
So the first time is going to be negative is when it intersects the axis at an intercept and it's going to dip below. This happens at , which is the 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.

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 . The number of ways to get an odd product can be counted like so: , which is , or . So, for one product to be odd the other to be even: (order matters). ~ Anonymous and Arctic_Bunny
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 (oddodd).
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 .

Solution 1 (Just Drop An Altitude)
It's crucial to draw a good diagram for this one. Since and , we get . Now we need to find 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 . Now expanding, simplifying, and dividing by the GCF, we get . This factors to . Since lengths cannot be negative, . Since , . So
(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 , and because the hypotenuse is 20 we get . Testing small numbers, we get that when and , 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 , , , ,and , respectively. Since lies on line , we know that . Furthermore, since , lies on the circle with diameter , so . Solving for and with these equations, we get the solutions and . We immediately discard the solution as should be negative. Thus, we conclude that .
Solution 4 (Trigonometry)
Let and Using Law of Sines on we getand LoS on yieldsDivide 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, and the answer is
(This solution is incomplete, can someone complete it pleaseLingjun) ok Latex edited by kc5170
We could use the famous mn rule in trigonometry in triangle ABC with Point E [Unable to write it here.Could 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. AB.AC = 120. Adding area of ABC and ACD Answer••360

Solution 1
First, substitute with . Then, the given equation becomes . Now consider only . This equals . Note that equals , 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 forms the sum of 17 different powers of 2. This gives us . But we must count also the term. Thus, Our answer is .
~seanyoon777
Solution 2
(This is similar to solution 1) Let . Then, . The LHS can be rewritten as . Plugging back in for , we have . When expanded, this will have terms. Therefore, our answer is .
Solution 3 (Intuitive)
Multiply both sides by to get
Notice that , since there is a on the LHS. However, now we have an extra term of on the right from . To cancel it, we let . The two 's now combine into a term of , so we let . And so on, 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 . So we let be .
As you can see, we will have to add 's at a time, then "wait" for the sum to automatically telescope for the next numbers, etc, until we get to . We only need to add 's between odd multiples of and even multiples. The largest even multiple of below is , so we will have to add a total of 's. However, we must not forget we let at the beginning, so our answer is .
Solution 4
Note that the expression is equal to something slightly lower than . Clearly, answer choices and make no sense because the lowest sum for terms is . just makes no sense. and are 1 apart, but because the expression is odd, it will have to contain , and because is bigger, the answer is .
~Lcz
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 in binary, we have
or
This means that each pair has terms of the form .Since there are of these pairs, there are a total of terms. Accounting for the term, which was not in the pair, we have a total of terms.

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 ; thus the three terms in the expression must be ,
if is an integer, then the three terms in the expression above must be ,
if is an integer, then the three terms in the expression above must be , and
if none of are integral, then the three terms in the expression above must be .
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
As mentioned above, the three terms in the expression are , so the sum is , which is divisible by . Therefore, the first case does not work (0 cases).
Case 2: divides and
As mentioned above, in this case the terms must be , which means the sum is , so the expression is not divisible by . Therefore, this is 1 case that works.
Case 3: divides
Because divides , the number of possibilities for is the same as the number of factors of .
= . So, the total number of factors of is .
However, we have to subtract , because the case does not work, as mentioned previously. This leaves 7 cases.
Case 4: divides
Because divides , the number of possibilities for is the same as the number of factors of .
= . So, the total number of factors of is .
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
Similar to Case 1, the value of the terms of the expression are . The sum is , which is divisible by 3, so this case does not work (0cases).
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 = .

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 to its original position. Notice that our set of rotations, , contains every multiple of except for . We can start with any two rotations in and there must be exactly one such that we can use the three rotations which ensures that . That way, the composition of rotations yields a full rotation. For example, if , then , so and the rotations yields a full rotation.
The only case in which this fails is when would have to equal . This happens when is already a full rotation, namely, 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 xaxis 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 xaxis would not map back to itself. The same applies to two reflections across the yaxis.
Therefore, we must use one reflection about the xaxis, one reflection about the yaxis, and one rotation. Since a reflection about the xaxis changes the sign of the y component, a reflection about the yaxis 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 and 8 can be achieved with any permutation of . This case can be done in ways.
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.

Solution 1
We know that , so we can write . Simplifying, we get . Similarly, we can write , or . Solving these two modular congruences, 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 . so again we add to . It turns out that does indeed satisfy the original conditions, so our answer is .
Solution 2 (bashing)
We are given that and . This tells us that is divisible by but not . It also tells us that is divisible by 60 but not 120. Starting, we find the least value of 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 , where , , , represent digits of the number (not to get confused with ). As given the problem, and . So we know that (last digit of ). That means that and . We can bash this after this. We just want to find all pairs of numbers such that is a multiple of 7 that is greater than a multiple of . Our equation for would be and our equation for would be , 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 which means that the sum of its digits is .
~ Baolan
Solution 4
The conditions of the problem reduce to the following. where and where . From these equations, we see that . Solving this diophantine equation gives us that , 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 conditions.Therefore, n is 1917 and =.happykeeper
Solution 6 (Reverse Euclidean Algorithm)
We are given that and By applying the Euclidean algorithm, but in reverse, we haveand
We now know that must be divisible by and so it is divisible by Therefore, for some integer We know that or else the first condition won't hold ( will be ) and or else the second condition won't hold ( will be ). Since gives us too small of an answer, then so the answer is

Solution 1
Consider the probability that rolling two dice gives a sum of , where . There are pairs that satisfy this, namely , 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 be the values of the three dice, which we will call , , and respectively. Consider the case when . If , then we do not need to reroll any dice. Otherwise, if we reroll one die, we can roll dice in the hope that we get the value that makes the sum of the three dice . This happens with probability . If we reroll two dice, we will roll and , and the probability of winning is , as stated above.
However, , so rolling one die is always better than rolling two dice if .
Now consider the case where . Rerolling one die will not help us win since the sum of the three dice will always be greater than . If we reroll two dice, the probability of winning is, once again, . To find the probability of winning if we reroll all three dice, we can let each dice have dot and find the number of ways to distribute the remaining dots. By Stars and Bars, there are ways to do this, making the probability of winning .
In order for rolling two dice to be more favorable than rolling three dice, .
Thus, rerolling two dice is optimal if and only if and . The possible triplets that satisfy these conditions, and the number of ways they can be permuted, are ways. ways. ways. ways. ways. ways. ways. ways. ways. ways.
There are ways in which rerolling two dice is optimal, out of possibilities, Therefore, the probability that Jason will reroll two dice is
Solution 2
We count the numerator. Jason will pick up no dice if he already has a 7 as a sum. We need to assume he does not have a 7 to begin with. If Jason decides to pick up all the dice to reroll, by Stars and Bars(or whatever...), there will be 2 bars and 4 stars(3 of them need to be guaranteed because a roll is at least 1) for a probability of . If Jason picks up 2 dice and leaves a die showing , he will need the other two to sum to . This happens with probabilityfor integers . If the roll is not 7, Jason will pick up exactly one die to reroll if there can remain two other dice with sum less than 7, since this will give him a chance which is a larger probability than all the cases unless he has a 7 to begin with. We haveWe count the underlined part's frequency for the numerator without upsetting the probability greater than it. Let be the roll we keep. We know is at most 3 since 4 would cause Jason to pick up all the dice. When , there are 3 choices for whether it is rolled 1st, 2nd, or 3rd, and in this case the other two rolls have to be at least 6(or he would have only picked up 1). This give ways. Similarly, gives because the 2 can be rolled in 3 places and the other two rolls are at least 5. gives . Summing together gives the numerator of 42. The denominator is , so we have