2021 AMC 10 B Fall

Solution 1:

We see that , and 4 each appear in the ones, tens, hundreds, and thousands digit exactly once. Since , we find that the sum is equal to Note that it is equally valid to manually add all four numbers together to get the answer.

Solution 2:

We have

Solution 1:

The line of symmetry divides the shaded figure into two congruent triangles, each with base 3 and height 2 . Therefore, the area of the shaded figure is

Solution 2:

To find the area of the shaded figure, we subtract the area of the smaller triangle (base 4 and height 2 ) from the area of the larger triangle (base 4 and height 5 ):

Solution 1:

We write the given expression as a single fraction: by cross multiplication. Then by factoring the numerator, we get The question is asking for the numerator, so our answer is , giving answer choice (E).

Solution 2:

Denote . Hence, We observe that and . Hence, . Therefore, . Therefore, the answer is (E) 4041

Solution 1:

At noon on a certain day, let and be the temperatures (in degrees) in Minneapolis and St. Louis, respectively. It follows that . At , we get We have two cases: 1. If , then . 2. If , then . Together, the product of all possible values of is .

Solution 2:

At noon on a certain day, the difference of temperatures in Minneapolis and St. Louis is degrees.

At , the difference of temperatures in Minneapolis and St. Louis is degrees. It follows that We continue with the casework in Solution 1 to get the answer .

Solution 1:

We have Therefore,

Solution 2:

The requested value is

Solution 1:

Let this positive integer be written as . The number of factors of this number is therefore , and this must equal 2021. The prime factorization of 2021 is , so and . To minimize this integer, we set and . Then this integer is . Now and so

Solution 2:

Recall that can be written as . Since we want the integer to have 2021 divisors, we must have it in the form , where and are prime numbers. Therefore, we want to be 3 and to be 2 . To make up the remaining , we multiply by , which is which is 16 . Therefore, we have

Solution 1:

The special fractions are We rewrite them in the simplest form: Note that two unlike fractions in the simplest form cannot sum to an integer. So, we only consider the fractions whose denominators appear more than once: For the set , two elements (not necessarily different) can sum to . For the set , two elements (not necessarily different) can sum to .

For the set , two elements (not necessarily different) can sum to 3 . Together, there are (C) 11 distinct integers that can be written as the sum of two, not necessarily different, special fractions:

Solution 2:

Let , so the special fraction is We can ignore the part and only focus on . The integers are , which are , respectively. We get from this group of numbers.

The halves are , which are , respectively. We get from this group of numbers. The quarters are , which are , respectively. We get 5 from this group of numbers. Note that 10 and 5 each appear twice. Therefore, the answer is (C) 11 .

We have Therefore, the greatest prime divisor of 16383 is 127 . The sum of its digits is .

Solution 1:

Let be the number of knights: then the number of red knights is and the number of blue knights is . Let be the fraction of blue knights that are magical - then is the fraction of red knights that are magical. Thus we can write the equation We want to find the fraction of red knights that are magical, which is

Solution 2:

We denote by the fraction of red knights who are magical. Hence, By solving this equation, we get . Therefore, the answer is (C) .

Solution 1:

Because Alice doesn't know who has the larger number, she doesn't have 1 . Because Alice says that she doesn't know who has the larger number, Bob knows that she doesn't have 1 . But Bob knows who has the larger number, this implies that Bob has the smallest possible number. Because Bob's number is prime, Bob's number is 2 . Thus, the perfect square is in the . The only perfect square is 225 . Thus, Alice's number is 25 . The sum of Alice's and Bob's number is . Thus the answer is .

Solution 2:

Denote by and the numbers drawn by Alice and Bob, respectively. Alice's sentence " 1 can't tell who has the larger number. implies . Bob's sentence "I know who has the larger number. implies . Their subsequent conversation that is prime implies . Then, Alice's next sentence "In that case, if I multiply your number by 100 and add my number, the result is a perfect square. implies is a perfect square. Hence, . Therefore, the answer is (A) 27 .

Solution 1:

This is the graph of the original Hexagon. After reflecting each minor arc over the sides of the hexagon it will look like this;

This bounded region is the same as the area of the hexagon minus the area of each of the reflect arcs. From the first diagram, the area of each arc is the area of the sector minus the area of the equilateral triangle, so each arc has an area of . There are 6 total arcs, so the total area of the arcs is . The area of the hexagon is , so the area of the bounded region is:

Solution 2:

Let the hexagon described be of area and let the circle's area be . Let the area we want to aim for be . Thus, we have that , or . By some formulas, and . Thus, or .

Solution 1:

It is obvious , and are symmetrical. We are going to solve the problem by Completing the Square. Because are integers, , and can only equal . So one variable must equal another, and the third variable is 1 different from those 2 equal variables. So the answer is .

Solution 2:

Plugging in every choice, we see that choice (D) works. We have , so Our answer is (D).

Solution 1:

Let's split the triangle down the middle and label it:

We see that by AA similarity. because cuts the side length of the square in half; similarly, . Let : then by side ratios, Now the height of the triangle is . By side ratios, The area of the triangle is

Solution 2:

By similarity, the height is and the base is . Thus the area is , or

We will use complementary counting to find the probability that the product is not divisible by . Then, we can find the probability that we want by subtracting this from . We split this into cases.

Case 1: The product is not divisible by We need every number to be odd, and since the chance we roll an odd number is , our probability is .

Case 2 : The product is divisible by , but not by . We need 5 numbers to be odd, and one to be divisible by 2 , but not by . There is a chance that an odd number is rolled, a chance that we roll a number satisfying the second condition (only 2 and 6 work), and 6 ways to choose the order in which the even number appears. Our probability is . Therefore, the probability the product is not divisible by is . Our answer is (C)

Solution 1:

Note that . Then, it follows that . Thus, . Define to be the length of side , then . Because is the altitude of the triangle, we can use the property that . Substituting the given lengths, we have Solving, gives and . We eliminate the possibilty of because . Thus, the side length of the square, by Pythagorean Theorem, is Thus, the area of the sqaure is . Thus, the answer is (D) 117

Solution 2:

As above, note that , which means that . In addition, note that is the altitude of a right triangle to its hypotenuse, so . Let the side length of the square be ; using similarity side ratios of to , we get Note that by the Pythagorean theorem, so we can use the expansion to produce two equations and two variables; We want , so we want to find . Subtracting the first equation from the second, we get Then

After the first swap, we do casework on the next swap.

Case 1: Silva swaps the two balls that were just swapped There is only one way for Silva to do this, and it leaves 5 balls occupying their original position.

Case 2: Silva swaps one ball that has just been swapped with one that hasn't swapped There are two ways for Silva to do this, and it leaves balls occupying their original positions.

Case 3 : Silva swaps two balls that have not been swapped There are two ways for Silva to do this, and it leaves balls occupying their original positions. Our answer is the average of all 5 possible swaps, so we get

Solution 1:

It is well known that the composition of 2 reflections, one after another, about two lines and , respectively, that meet at an angle is a rotation by around the intersection of and . Now, we note that is a 90 degree rotation clockwise of about the origin, which is also where and intersect. So is a 45 degree rotation of about the origin clockwise. To rotate degrees clockwise, we build a square with adjacent vertices and . The other two vertices are at and . The center of the square is at , which is the midpoint of and . The line passes through the origin and the center of the square we built, namely at and . Thus the line is . The answer is (D) .

Solution 2:

We know that the equation of line is . This means that is reflected over the line . This means that the line with and is perpendicular to , so it has slope . Then the equation of this perpendicular line is , and plugging in for and yields .

The midpoint of and lies at the intersection of and . Solving, we get the -value of the intersection is and the -value is . Let the x-value of be then by the midpoint formula, . We can find the -value of the same way, so .

Now we have to reflect over to get to . The midpoint of and will lie on , and this midpoint is, by the midpoint formula, must satisfy this point, so Now the equation of line is