2020 AMC12A 真题及答案
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参考答案见文末（仅供参考）
Problem 1
What is the value in simplest form of the following expression?
Problem 2
What is the value of the following expression?
Problem 3
The ratio of to is , the ratio of to is , and the ratio of to is . What is the ratio of to ?
Problem 4
The acute angles of a right triangle are and , where and both and are prime numbers. What is the least possible value of ?
Problem 5
Teams and are playing in a basketball league where each game results in a win for one team and a loss for the other team. Team has won of its games and team has won of its games. Also, team has won more games and lost more games than team How many games has team played?
Problem 6
For all integers the value ofis always which of the following?
Problem 7
Two nonhorizontal, non vertical lines in the coordinate plane intersect to form a angle. One line has slope equal to times the slope of the other line. What is the greatest possible value of the product of the slopes of the two lines?
Problem 8
How many ordered pairs of integers satisfy the equation
Problem 9
A threequarter sector of a circle of radius inches together with its interior can be rolled up to form the lateral surface area of a right circular cone by taping together along the two radii shown. What is the volume of the cone in cubic inches?
Problem 10
In unit square the inscribed circle intersects at and intersects at a point different from What is
Problem 11
As shown in the figure below, six semicircles lie in the interior of a regular hexagon with side length so that the diameters of the semicircles coincide with the sides of the hexagon. What is the area of the shaded region—inside the hexagon but outside all of the semicircles?
Problem 12
Let be a diameter in a circle of radius Let be a chord in the circle that intersects at a point such that and What is
Problem 13
Which of the following is the value of
Problem 14
Bela and Jenn play the following game on the closed interval of the real number line, where is a fixed integer greater than . They take turns playing, with Bela going first. At his first turn, Bela chooses any real number in the interval . Thereafter, the player whose turn it is chooses a real number that is more than one unit away from all numbers previously chosen by either player. A player unable to choose such a number loses. Using optimal strategy, which player will win the game?
Problem 15
There are 10 people standing equally spaced around a circle. Each person knows exactly 3 of the other 9 people: the 2 people standing next to her or him, as well as the person directly across the circle. How many ways are there for the 10 people to split up into 5 pairs so that the members of each pair know each other?
Problem 16
An urn contains one red ball and one blue ball. A box of extra red and blue balls lie nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the urn contains six balls. What is the probability that the urn contains three balls of each color?
Problem 17
How many polynomials of the form , where , , , and are real numbers, have the property that whenever is a root, so is ? (Note that )
Problem 18
In square , points and lie on and , respectively, so that Points and lie on and , respectively, and points and lie on so that and . See the figure below. Triangle , quadrilateral , quadrilateral , and pentagon each has area What is ?
Problem 19
Square in the coordinate plane has vertices at the points and Consider the following four transformations: a rotation of counterclockwise around the origin; a rotation of clockwise around the origin; a reflection across the axis; and a reflection across the axis.
Each of these transformations maps the squares onto itself, but the positions of the labeled vertices will change. For example, applying and then would send the vertex at to and would send the vertex at to itself. How many sequences of transformations chosen from will send all of the labeled vertices back to their original positions? (For example, is one sequence of transformations that will send the vertices back to their original positions.)
Problem 20
Two different cubes of the same size are to be painted, with the color of each face being chosen independently and at random to be either black or white. What is the probability that after they are painted, the cubes can be rotated to be identical in appearance?
Problem 21
How many positive integers satisfy(Recall that is the greatest integer not exceeding .)
Problem 22
What is the maximum value of for real values of
Problem 23
How many integers are there such that whenever are complex numbers such that
then the numbers are equally spaced on the unit circle in the complex plane?
Problem 24
Let denote the number of ways of writing the positive integer as a productwhere , the are integers strictly greater than , and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are counted as distinct). For example, the number can be written as , , and , so . What is ?
Problem 25
For each real number with , let numbers and be chosen independently at random from the intervals and , respectively, and let be the probability that
What is the maximum value of
 =
Note: This comes from the fact that the sum of the first odds is .  Using difference of squares to factor the left term, we getCancelling all the terms, we get as the answer.

Solution 1
WLOG, let and .
Since the ratio of to is , we can substitute in the value of to get .
The ratio of to is , so .
The ratio of to is then so our answer is ~quacker88
Solution 2
We need to somehow link all three of the ratios together. We can start by connecting the last two ratios together by multiplying the last ratio by two.
, and since , we can link them together to get .
Finally, since , we can link this again to get: , so

Solution 1
Since the three angles of a triangle add up to and one of the angles is because it's a right triangle, .
The greatest prime number less than is . If , then , which is not prime.
The next greatest prime number less than is . If , then , which IS prime, so we have our answer
Solution 2
Looking at the answer choices, only and are coprime to . Testing , the smaller angle, makes the other angle which is prime, therefore our answer is

Solution
Let's first find how many possibilities there would be if they were all distinguishable, then divide out the ones we overcounted.
There are ways to order objects. However, since there's ways to switch the yellow tiles around without changing anything (since they're indistinguishable) and ways to order the green tiles, we have to divide out these possibilities.
~quacker88
Solution
We can repeat chooses extensively to find the answer. There are choose ways to arrange the brown tiles which is . Then from the remaining tiles there are choose ways to arrange the red tiles. And now from the remaining two tiles and two slots we can see there are two ways to arrange the purple and brown tiles, giving us an answer of  noahdavid
 Any even multiple of is a multiple of , so we need to find multiples of that are perfect squares and less than . Any solution that we want will be in the form , where is a positive integer. The smallest possible value is at , and the largest is at (where the expression equals ). Therefore, there are a total of possible numbers.PCChess
 Any even multiple of is a multiple of , so we need to find multiples of that are perfect squares and less than . Any solution that we want will be in the form , where is a positive integer. The smallest possible value is at , and the largest is at (where the expression equals ). Therefore, there are a total of possible numbers.PCChess

Solution 1
There are options here:
1. is the right angle.
It's clear that there are points that fit this, one that's directly to the right of and one that's directly to the left. We don't need to find the length, we just need to know that it is possible, which it is.
2. is the right angle.
Using the exact same reasoning, there are also solutions for this one.
3. The new point is the right angle.
(Diagram temporarily removed due to asymptote error)
The diagram looks something like this. We know that the altitude to base must be since the area is . From here, we must see if there are valid triangles that satisfy the necessary requirements.
First of all, because of the area.
Next, from the Pythagorean Theorem.
From here, we must look to see if there are valid solutions. There are multiple ways to do this:
We know that the minimum value of is when . In this case, the equation becomes , which is LESS than . . The equation becomes , which is obviously greater than . We can conclude that there are values for and in between that satisfy the Pythagorean Theorem.
And since , the triangle is not isoceles, meaning we could reflect it over and/or the line perpendicular to for a total of triangles this case.
Solution 2
Note that line segment can either be the shorter leg, longer leg or the hypotenuse. If it is the shorter leg, there are two possible points for that can satisfy the requirements  that being above or below . As such, there are ways for this case. Similarly, one can find that there are also ways for point to lie if is the longer leg. If it is a hypotenuse, then there are possible points because the arrangement of the shorter and longer legs can be switched, and can be either above or below the line segment. Therefore, the answer is .

Solution
Rearranging the terms and and completing the square for yields the result . Then, notice that can only be , and because any value of that is greater than 1 will cause the term to be less than , which is impossible as must be real. Therefore, plugging in the above values for gives the ordered pairs , , , and gives a total of ordered pairs.
Solution 2
Bringing all of the terms to the LHS, we see a quadratic equationin terms of . Applying the quadratic formula, we getIn order for to be real, which it must be given the stipulation that we are seeking integral answers, we know that the discriminant, must be nonnegative. Therefore,Here, we see that we must split the inequality into a compound, resulting in .
The only integers that satisfy this are . Plugging these values back into the quadratic equation, we see that both produce a discriminant of , meaning that there is only 1 solution for . If , then the discriminant is nonzero, therefore resulting in two solutions for .
Thus, the answer is .
~Tiblis
Solution 3, x first
Set it up as a quadratic in terms of y:Then the discriminant isThis will clearly only yield real solutions when , because it is always positive. Then . Checking each one: and are the same when raised to the 2020th power:This has only has solutions , so are solutions. Next, if :Which has 2 solutions, so and
These are the only 4 solutions, so
Solution 4, y first
Move the term to the other side to get . Because for all , then . If or , the right side is and therefore . When , the right side become , therefore . Our solutions are , , , . There are solutions, so the answer is  wwt7535

Solution
Notice that when the cone is created, the radius of the circle will become the slant height of the cone and the intact circumference of the circle will become the circumference of the base of the cone.
We can calculate that the intact circumference of the circle is . Since that is also equal to the circumference of the cone, the radius of the cone is . We also have that the slant height of the cone is . Therefore, we use the Pythagorean Theorem to calculate that the height of the cone is . The volume of the cone is PCChess
Solution 2 (Last Resort/Cheap)
Using a ruler, measure a circle of radius 4 and cut out the circle and then the quarter missing. Then, fold it into a cone and measure the diameter to be 6 cm . You can form a right triangle with sides 3, 4, and then through the Pythagorean theorem the height is found to be . The volume of a cone is . Plugging in we find
 We don't care about which books Harold selects. We just care that Betty picks books from Harold's list and that aren't on Harold's list.The total amount of combinations of books that Betty can select is .There are ways for Betty to choose of the books that are on Harold's list.From the remaining books that aren't on Harold's list, there are ways to choose of them. ~quacker88

Solution 1
Now we do some estimation. Notice that , which means that is a little more than . Multiplying it with , we get that the denominator is about . Notice that when we divide by an digit number, there are zeros before the first nonzero digit. This means that when we divide by the digit integer , there are zeros in the initial string after the decimal point. PCChess
Solution 2
First rewrite as . Then, we know that when we write this in decimal form, there will be 40 digits after the decimal point. Therefore, we just have to find the number of digits in .
and memming (alternatively use the fact that ), digits.
Our answer is .
Solution 3 (Brute Force)
Just as in Solution we rewrite as We then calculate entirely by hand, first doing then multiplying that product by itself, resulting in Because this is digits, after dividing this number by fourteen times, the decimal point is before the Dividing the number again by twentysix more times allows a string of zeroes to be formed. OreoChocolate
Solution 4 (Smarter Brute Force)
Just as in Solutions and we rewrite as We can then look at the number of digits in powers of . , , , , , , and so on. We notice after a few iterations that every power of five with an exponent of , the number of digits doesn't increase. This means should have digits since there are numbers which are from to , or digits total. This means our expression can be written as , where is in the range . Canceling gives , or zeroes before the since the number should start on where the one would be in . ~aop2014
Solution 5 (Logarithms)
~phoenixfire

Solution 1
You can find that every four moves both coordinates decrease by 2. Therefore, both coordinates need to decrease by two 505 times. You subtract, giving you the answer of ~happykeeper
See Also
2015 AMC 10B Problem 24

Solution 1
Let point A be a vertex of the regular hexagon, let point B be the midpoint of the line connecting point A and a neighboring vertex, and let point C be the second intersection of the two semicircles that pass through point A. Then, , since B is the center of the semicircle with radius 1 that C lies on, , since B is the center of the semicircle with radius 1 that A lies on, and , as a regular hexagon has angles of 120, and is half of any angle in this hexagon. Now, using the sine law, , so . Since the angles in a triangle sum to 180, is also 60. Therefore, is an equilateral triangle with side lengths of 1.
Since the area of a regular hexagon can be found with the formula , where is the side length of the hexagon, the area of this hexagon is . Since the area of an equilateral triangle can be found with the formula , where is the side length of the equilateral triangle, the area of an equilateral triangle with side lengths of 1 is . Since the area of a circle can be found with the formula , the area of a sixth of a circle with radius 1 is . In each sixth of the hexagon, there are two equilateral triangles colored white, each with an area of , and one sixth of a circle with radius 1 colored white, with an area of . The rest of the sixth is colored gray. Therefore, the total area that is colored white in each sixth of the hexagon is , which equals , and the total area colored white is , which equals . Since the area colored gray equals the total area of the hexagon minus the area colored white, the area colored gray is , which equals .
Solution 2
First, subdivide the hexagon into 24 equilateral triangles with side length 1:Now note that the entire shaded region is just 6 times this part:The entire rhombus is just 2 equilatrial triangles with side lengths of 1, so it has an area of:The arc that is not included has an area of:Hence, the area of the shaded region in that section isFor a final area of:

Solution 1
After erasing every third digit, the list becomes repeated. After erasing every fourth digit from this list, the list becomes repeated. Finally, after erasing every fifth digit from this list, the list becomes repeated. Since this list repeats every digits and since are respectively in we have that the th, th, and st digits are the rd, th, and th digits respectively. It follows that the answer is

Solution
Notice that to use the optimal strategy to win the game, Bela must select the middle number in the range and then mirror whatever number Jenn selects. Therefore, if Jenn can select a number within the range, so can Bela. Jenn will always be the first person to run out of a number to choose, so the answer is .
Solution 2 (Guessing)
First of all, realize that the value of should have no effect on the strategy at all. This is because they can choose real numbers, not integers, so even if is odd, for example, they can still go halfway. Similarly, there is no reason the strategy would change when .
So we are left with (A) and (B). From here it is best to try out random numbers and try to find the strategy that will let Bela win, but if you can't find it, realize that it is more likely the answer is since Bela has the first move and thus has more control.
 Let us use casework on the number of diagonals.Case 1: diagonals There are ways: either pairs with , pairs with , and so on or pairs with , pairs with , etc.Case 2: diagonal There are possible diagonals to draw (everyone else pairs with the person next to them.Note that there cannot be 2 diagonals.Case 3: diagonalsNote that there cannot be a case with 4 diagonals because then there would have to be 5 diagonals for the two remaining people, thus a contradiction.Case 4: diagonals There is way to do this.Thus, in total there are possible ways.

Solution
Let denote that George selects a red ball and that he selects a blue one. Now, in order to get balls of each color, he needs more of both and .
There are 6 cases: (we can confirm that there are only since ). However we can clump , , and together since they are equivalent by symmetry.
and
Let's find the probability that he picks the balls in the order of .
The probability that the first ball he picks is red is .
Now there are reds and blue in the urn. The probability that he picks red again is now .
There are reds and blue now. The probability that he picks a blue is .
Finally, there are reds and blues. The probability that he picks a blue is .
So the probability that the case happens is . However, since the case is the exact same by symmetry, case 1 has a probability of chance of happening.
and
Let's find the probability that he picks the balls in the order of .
The probability that the first ball he picks is red is .
Now there are reds and blue in the urn. The probability that he picks blue is .
There are reds and blues now. The probability that he picks a red is .
Finally, there are reds and blues. The probability that he picks a blue is .
So the probability that the case happens is . However, since the case is the exact same by symmetry, case 2 has a probability of chance of happening.
and
Let's find the probability that he picks the balls in the order of .
The probability that the first ball he picks is red is .
Now there are reds and blue in the urn. The probability that he picks blue is .
There are reds and blues now. The probability that he picks a blue is .
Finally, there are reds and blues. The probability that he picks a red is .
So the probability that the case happens is . However, since the case is the exact same by symmetry, case 3 has a probability of chance of happening.
Adding up the cases, we have ~quacker88
Solution 2
We know that we need to find the probability of adding 2 red and 2 blue balls in some order. There are 6 ways to do this, since there are ways to arrange in some order. We will show that the probability for each of these 6 ways is the same.
We first note that the denominators should be counted by the same number. This number is . This is because 2, 3, 4, and 5 represent how many choices there are for the four steps. No matter what the step involves numbers to choose from.
The numerators are the number of successful operations. No matter the order, the first time a red is added will come from 1 choice and the second time will come from 2 choices, since that is how many reds are in the urn originally. The same goes for the blue ones. The numerator must equal .
Therefore, the probability for each of the orderings of is . There are 6 of these, so the total probability is .
Solution 3
First, notice that when George chooses a ball he just adds another ball of the same color. On George's first move, he either chooses the red or the blue with a chance each. We can assume he chooses Red(chance ), and then multiply the final answer by two for symmetry. Now, there are two red balls and one blue ball in the urn. Then, he can either choose another Red(chance ), in which case he must choose two blues to get three of each, with probability or a blue for two blue and two red in the urn, with chance . If he chooses blue next, he can either choose a red then a blue, or a blue then a red. Each of these has a for total of . The total probability that he ends up with three red and three blue is . ~aop2014
Solution 4
Let the probability that the urn ends up with more red balls be denoted . Since this is equal to the probability there are more blue balls, the probability there are equal amounts is . the probability no more blues are chosen plus the probability only 1 more blue is chosen. The first case, .
The second case, . Thus, the answer is .
~JHawk0224
Solution 5
By conditional probability after 4 rounds we have 5 cases: RRRBBB, RRRRBB, RRBBBB, RRRRRB and RBBBBB. Thus the probability is . Put .
~FANYUCHEN20020715
Edited by Kinglogic
Solution 6
Here X stands for R or B, and Y for the remaining color. After 3 rounds one can either have a 4+1 configuration (XXXXY), or 3+2 configuration (XXXYY). The probability of getting to XXXYYY from XXXYY is . Observe that the probability of arriving to 4+1 configuration is( to get from XXY to XXXY, to get from XXXY to XXXXY). Thus the probability of arriving to 3+2 configuration is also , and the answer is
Solution 7
We can try to use dynamic programming to solve this problem. (Informatics Olympiad hahaha)
We let be the probability that we end up with red balls and blue balls. Notice that there are only two ways that we can end up with red balls and blue balls: one is by fetching a red ball from the urn when we have red balls and blue balls and the other is by fetching a blue ball from the urn when we have red balls and blue balls.
Then we have
Then we start can with and try to compute .
The answer is .
(Solution by CircleOO)

Solution 1
We're looking for the amount of ways we can get cards from a deck of , which is represented by .
We need to get rid of the multiples of , which will subsequently get rid of the multiples of (if we didn't, the zeroes would mess with the equation since you can't divide by 0)
, , leaves us with 17.
Converting these into, we have
~quacker88
Solution 2
Since this number is divisible by but not , the last digits must be divisible by but the last digits cannot be divisible by . This narrows the options down to and .
Also, the number cannot be divisible by . Adding up the digits, we get . If , then the expression equals , a multiple of . This would mean that the entire number would be divisible by , which is not what we want. Therefore, the only option is PCChess
Solution 3
It is not hard to check that divides the number,As , using we have . Thus , implying so the answer is .
Solution 4
As mentioned above,
We can divide both sides of by 10 to obtainwhich means is simply the units digit of the lefthand side. This value is  Split into 4 regions:1. The rectangular prism itself2. The extensions of the faces of 3. The quarter cylinders at each edge of 4. The oneeighth spheres at each corner of Region 1: The volume of is 12, so Region 2: The volume is equal to the surface area of times . The surface area can be computed to be , so .Region 3: The volume of each quarter cylinder is equal to . The sum of all such cylinders must equal times the sum of the edge lengths. This can be computed as , so the sum of the volumes of the quarter cylinders is , so Region 4: There is an eighth of a sphere of radius at each corner. Since there are 8 corners, these add up to one full sphere of radius . The volume of this sphere is , so .
Using these values,

Solution
Since the total area is , the side length of square is . We see that since triangle is a right isosceles triangle with area 1, we can determine sides and both to be . Now, consider extending and until they intersect. Let the point of intersection be . We note that is also a right isosceles triangle with side and find it's area to be . Now, we notice that is also a right isosceles triangle and find it's area to be . This is also equal to or . Since we are looking for , we want two times this. That gives .~TLiu
Solution 2
Since this is a geometry problem involving sides, and we know that is , we can use our ruler and find the ratio between and . Measuring(on the booklet), we get that is about inches and is about inches. Thus, we can then multiply the length of by the ratio of of which we then get We take the square of that and get and the closest answer to that is . ~Celloboy (Note that this is just a strategy I happened to use that worked. Do not press your luck with this strategy, for it was a lucky guess)
Solution 3
Draw the auxiliary line . Denote by the point it intersects with , and by the point it intersects with . Last, denote by the segment , and by the segment . We will find two equations for and , and then solve for .
Since the overall area of is , and . In addition, the area of .
The two equations for and are then:
Length of :
Area of CMIF: .
Substituting the first into the second, yields
Solving for gives ~DrB
Solution 4
Plot a point such that and are collinear and extend line to point such that forms a square. Extend line to meet line and point is the intersection of the two. The area of this square is equivalent to . We see that the area of square is , meaning each side is of length 2. The area of the pentagon is . Length , thus . Triangle is isosceles, and the area of this triangle is . Adding these two areas, we get
Solution 5 (HARD Calculation)
We can easily observe that the area of square is 4 and its side length is 2 since all four regions that build up the square has area 1. Extend and let the intersection with be . Connect , and let the intersection of and be . Notice that since the area of triangle is 1 and , , therefore . Let , then . Also notice that , thus . Now use the condition that the area of quadrilateral is 1, we can set up the following equation: We solve the equation and yield . Now notice that . Hence . HarryW

Solution
Let . We are now looking for the remainder of .
We could proceed with polynomial division, but the denominator looks awfully similar to the Sophie Germain Identity, which states that
Let's use the identity, with and , so we have
Rearranging, we can see that this is exactly what we need:
So
Since the first half divides cleanly as shown earlier, the remainder must be ~quacker88
Solution 2
Similar to Solution 1, let . It suffices to find remainder of . Dividing polynomials results in a remainder of .
MAA Original Solution
Thus, we see that the remainder is surely

Solution
Let (+) denote counterclockwise/starting orientation and () denote clockwise orientation. Let 1,2,3, and 4 denote which quadrant A is in.
Realize that from any odd quadrant and any orientation, the 4 transformations result in some permutation of .
The same goes that from any even quadrant and any orientation, the 4 transformations result in some permutation of .
We start our first 19 moves by doing whatever we want, 4 choices each time. Since 19 is odd, we must end up on an even quadrant.
As said above, we know that exactly one of the four transformations will give us , and we must use that transformation.
Thus
Solution 2
Hopefully, someone will think of a better one, but here is an indirect answer, use only if you are really desperate. moves can be made, and each move have choices, so a total of moves. First, after the moves, Point A can only be in first quadrant or third quadrant . Only the one in the first quadrant works, so divide by . Now, C must be in the opposite quadrant as A. B can be either in the second () or fourth quadrant () , but we want it to be in the second quadrant, so divide by again. Now as A and B satisfy the conditions, C and D will also be at their original spot. . The answer is ~Kinglogic
Solution 3
The total number of sequence is .
Note that there can only be even number of reflections since they result in the same anticlockwise orientation of the verices . Therefore, the probability of having the same anticlockwise orientation with the original arrangement after the transformation is .
Next, even number of reflections mean that there must be even number of rotations since their sum is even. Even rotations result only in the original position or rotation of it.
Since rotation and rotation cancels each other out, the difference between the numbers of them define the final position. The probability of the transformation returning the vertices to the orginal position given that there are even number of rotations is equivalent to the probability that
when or when which is again, .
Therefore, ~joshuamh111
Solution 4
Notice that any pair of two of these transformations either swaps the x and ycoordinates, negates the x and ycoordinates, swaps and negates the x and ycoordinates, or leaves the original unchanged. Furthermore, notice that for each of these results, if we apply another pair of transformations, one of these results will happen again, and with equal probability. Therefore, no matter what state after we apply the first pairs of transformations, there is a chance the last pair of transformations will return the figure to its original position. Therefore, the answer is

Solution
First notice that the graphs of and intersect at 2 points. Then, notice that must be an integer. This means that n is congruent to .
For the first intersection, testing the first few values of (adding to each time and noticing the left side increases by each time) yields and . timating from the graph can narrow down the other cases, being , d . This results in a total of 6 cases, for an answer of .
~DrJoyo
Solution 2 (Graphing)
One intuitive approach to the question is graphing. Obviously, you should know what the graph of the square root function is, and if any function is floored (meaning it is taken to the greatest integer less than a value), a stairlike figure should appear. The other function is simply a line with a slope of . If you precisely draw out the two regions of the graph where the derivative of the square function nears the derivative of the linear function, you can now deduce that values of intersection lay closer to the left side of the stair, and values lay closer to the right side of the stair.
With meticulous graphing, you can realize that the answer is .
A indepth graph with intersection points is linked below. https://www.desmos.com/calculator/e5wk9adbuk
Solution 3
 Not a reliable or indepth solution (for the guess and check students)
We can first consider the equation without a floor function:
Multiplying both sides by 70 and then squaring:
Moving all terms to the left:
Now we can use wishful thinking to determine the factors:
This means that for and , the equation will hold without the floor function.
Now we can simply check the multiples of 70 around 400 and 2500 in the original equation:
For , left hand side but so right hand side
For , left hand side and right hand side
For , left hand side and right hand side
For , left hand side but so right hand side
Now we move to
For , left hand side and so right hand side
For , left hand side and so right hand side
For , left hand side and so right hand side
For , left hand side but so right hand side
For , left hand side and right hand side
For , left hand side but so right hand side
Therefore we have 6 total solutions,
Solution 4
This is my first solution here, so please forgive me for any errors.
We are given that
must be an integer, which means that is divisible by . As , this means that , so we can write for .
Therefore,
Also, we can say that and
Squaring the second inequality, we get .
Similarly solving the first inequality gives us or
is slightly larger than , so instead, we can say or .
Combining this with , we get are all solutions for that give a valid solution for , meaning that our answer is .
Solution By Qqqwerw
Solution 5
We start with the given equationFrom there, we can start with the general inequality that . This means thatSolving each inequality separately gives us two inequalities:Simplifying and approximating decimals yields 2 solutions for one inequality and 4 for the other. Hence .

Solution
Note that . Since there are at most six not necessarily distinct factors multiplying to , we have six cases: Now we look at each of the six cases.
: We see that there is way, merely .
: This way, we have the in one slot and in another, and symmetry. The four other 's leave us with ways and symmetry doubles us so we have .
: We have as our baseline. We need to multiply by in places, and see that we can split the remaining three powers of in a manner that is , or . A split has ways of happening ( and symmetry; and symmetry), a split has ways of happening (due to all being distinct) and a split has ways of happening ( and symmetry) so in this case we have ways.
: We have as our baseline, and for the two other 's, we have a or split. The former grants us ways ( and symmetry and and symmetry) and the latter grants us also ways ( and symmetry and and symmetry) for a total of ways.
: We have as our baseline and one place to put the last two: on another two or on the three. On the three gives us ways due to symmetry and on another two gives us ways due to symmetry. Thus, we have ways.
: We have and symmetry and no more twos to multiply, so by symmetry, we have ways.
Thus, adding, we have .
~kevinmathz
Solution 2
As before, note that , and we need to consider 6 different cases, one for each possible value of , the number of factors in our factorization. However, instead of looking at each individually, find a general form for the number of possible factorizations with factors. First, the factorization needs to contain one factor that is itself a multiple of , and there are to choose from, and the rest must contain at least one factor of . Next, consider the remaining factors of left to assign to the factors. Using stars and bars, the number of ways to do this isThis makes possibilities for each k.
To obtain the total number of factorizations, add all possible values for k:.
Solution 3
Begin by examining . can take on any value that is a factor of except . For each choice of , the resulting must have a product of . This means the number of ways the rest , can be written by the scheme stated in the problem for each is equal to , since the product of is counted as one valid product if and only if , the product has the properties that factors are greater than , and differently ordered products are counted separately.
For example, say the first factor is . Then, the remaining numbers must multiply to , so the number of ways the product can be written beginning with is . To add up all the number of solutions for every possible starting factor, must be calculated and summed for all possible , except and , since a single is not counted according to the problem statement. The however, is counted, but only results in possibility, the first and only factor being . This means
.
Instead of calculating D for the larger factors first, reduce , , and into sums of where to ease calculation. Following the recursive definition sums of where c takes on every divisor of n except for 1 and itself, the sum simplifies to
+
, so the sum further simplifies to
, after combining terms. From quick casework,
and . Substituting these values into the expression above,
.
~monmath a.k.a Fmirza
Solution 4
Note that , and that of a perfect power of a prime is relatively easy to calculate. Also note that you can find from by simply totaling the number of ways there are to insert a into a set of numbers that multiply to .
First, calculate . Since , all you have to do was find the number of ways to divide the 's into groups, such that each group has at least one . By stars and bars, this results in way with five terms, ways with four terms, ways with three terms, ways with two terms, and way with one term. (The total, , is not needed for the remaining calculations.)
Then, to get , in each possible sequence, insert a somewhere in it, either by placing it somewhere next to the original numbers (in one of ways, where is the number of terms in the sequence), or by multiplying one of the numbers by (in one of ways). There are ways to do this with one term, with two, with three, with four, and with five.
The resulting number of possible sequences is .