amc math - Jan 14, 07:22

Solution

Correct: E

We use Fermat's Little Theorem, which states that if p is a prime number, then for any integer a not divisible by p, a^(p-1) is congruent to 1 (mod p). Here, p = 23 and a = 3. Since 23 is a prime number and 3 is not divisible by 23, we have 3^(23-1) = 3^22 is congruent to 1 (mod 23). Now we need to find the remainder of the exponent 2023 when divided by 22: 2023 = 22 * 92 + 9. So, 3^2023 = 3^(22*92 + 9) = (3^22)^92 * 3^9 (mod 23). Since 3^22 is congruent to 1 (mod 23), we have: (3^22)^92 * 3^9 is congruent to 1^92 * 3^9 (mod 23). This simplifies to 3^9 (mod 23). Now we calculate powers of 3 modulo 23: 3^1 = 3 3^2 = 9 3^3 = 27 which is congruent to 4 (mod 23) 3^4 = 3^3 * 3 is congruent to 4 * 3 = 12 (mod 23) 3^5 = 12 * 3 = 36 which is congruent to 13 (mod 23) 3^6 = 13 * 3 = 39 which is congruent to 16 (mod 23) 3^7 = 16 * 3 = 48 which is congruent to 2 (mod 23) 3^8 = 2 * 3 = 6 (mod 23) 3^9 = 6 * 3 = 18 (mod 23). Thus, the remainder when 3^2023 is divided by 23 is 18.

Solution

Correct: E

From Vieta's formulas for the polynomial P(x) = x^3 - 3x^2 + 2x - 1, we have: 1. Sum of the roots: a + b + c = -(-3)/1 = 3 2. Sum of the products of the roots taken two at a time: ab + bc + ca = 2/1 = 2 3. Product of the roots: abc = -(-1)/1 = 1 We want to find the value of (a+b)(b+c)(c+a). From (1), we can express each term in the product: a + b = 3 - c b + c = 3 - a c + a = 3 - b Substitute these into the expression: (a+b)(b+c)(c+a) = (3-c)(3-a)(3-b). This expression is related to the polynomial P(x). Recall that if a, b, c are the roots of P(x), then P(x) can be written as (x-a)(x-b)(x-c) (assuming the leading coefficient is 1, which it is in this case). So, P(x) = (x-a)(x-b)(x-c). The expression (3-c)(3-a)(3-b) is exactly P(3). Let's calculate P(3): P(3) = 3^3 - 3(3^2) + 2(3) - 1 P(3) = 27 - 3(9) + 6 - 1 P(3) = 27 - 27 + 6 - 1 P(3) = 5. Thus, the value of (a+b)(b+c)(c+a) is 5.

Solution

Correct: C

Let's set up a coordinate system. Let A = (0,0), B = (1,0), C = (1,1), D = (0,1). Since M is the midpoint of AB, M = ( (0+1)/2, (0+0)/2 ) = (1/2, 0). Since N is the midpoint of BC, N = ( (1+1)/2, (0+1)/2 ) = (1, 1/2). Next, we find the equations of lines CM and DN. Line CM passes through C(1,1) and M(1/2,0). Slope of CM = (1-0) / (1-1/2) = 1 / (1/2) = 2. Equation of CM: y - 0 = 2(x - 1/2) => y = 2x - 1. Line DN passes through D(0,1) and N(1,1/2). Slope of DN = (1/2-1) / (1-0) = (-1/2) / 1 = -1/2. Equation of DN: y - 1 = -1/2 (x - 0) => y = -1/2 x + 1. Point P is the intersection of CM and DN. We set the y-values equal: 2x - 1 = -1/2 x + 1 2x + 1/2 x = 1 + 1 5/2 x = 2 x = 4/5. Now find the y-coordinate of P using either equation: y = 2(4/5) - 1 = 8/5 - 5/5 = 3/5. So, the coordinates of point P are (4/5, 3/5). We need to find the area of triangle APD. The vertices are A(0,0), P(4/5, 3/5), D(0,1). We can choose AD as the base of the triangle. The length of base AD is the distance between (0,0) and (0,1), which is 1. Since AD lies along the y-axis, the height of the triangle with respect to this base is the perpendicular distance from P to the y-axis, which is the x-coordinate of P. Height = 4/5. Area of triangle APD = 1/2 * base * height = 1/2 * 1 * (4/5) = 2/5.

Solution

Correct: C

A positive integer has exactly three divisors if and only if it is the square of a prime number. Let the integer be n = p^2, where p is a prime number. We are looking for integers n such that 100 < n < 200. So, we need to find prime numbers p such that 100 < p^2 < 200. Taking the square root of the inequality: sqrt(100) < p < sqrt(200) 10 < p < sqrt(200). We know that 14^2 = 196, and 15^2 = 225. So, sqrt(200) is approximately 14.14. Therefore, we are looking for prime numbers p such that 10 < p < 14.14. The prime numbers in this range are: - p = 11 (since 11 is prime and 10 < 11 < 14.14) - p = 13 (since 13 is prime and 10 < 13 < 14.14) Now we find the squares of these primes: - For p = 11, n = 11^2 = 121. The divisors of 121 are 1, 11, 121 (exactly three divisors). 100 < 121 < 200. - For p = 13, n = 13^2 = 169. The divisors of 169 are 1, 13, 169 (exactly three divisors). 100 < 169 < 200. For the next prime, p = 17, n = 17^2 = 289, which is greater than 200. Thus, there are two such positive integers: 121 and 169.

Solution

Correct: B

We are given the functional equation f(x+y) = f(x) + f(y) + xy and f(1) = 1. We need to find f(3). Let's find f(2) first: Set x = 1 and y = 1: f(1+1) = f(1) + f(1) + (1)(1) f(2) = f(1) + f(1) + 1 Since f(1) = 1: f(2) = 1 + 1 + 1 = 3. Now, let's find f(3): We can set x = 2 and y = 1: f(2+1) = f(2) + f(1) + (2)(1) f(3) = f(2) + f(1) + 2 Substitute the values we found for f(2) and f(1): f(3) = 3 + 1 + 2 f(3) = 6. Alternatively, we could set x=1, y=2: f(1+2) = f(1) + f(2) + (1)(2) f(3) = f(1) + f(2) + 2 f(3) = 1 + 3 + 2 = 6. (Consistent) Another approach could be to find the general form of f(x). If g(x) = f(x) - x^2/2, then g(x+y) = f(x+y) - (x+y)^2/2 = f(x)+f(y)+xy - (x^2+2xy+y^2)/2 = f(x)+f(y)+xy - x^2/2 - xy - y^2/2 = (f(x)-x^2/2) + (f(y)-y^2/2) = g(x)+g(y). This is Cauchy's functional equation. If g(x)=cx for rational x, then f(x) = cx + x^2/2. Given f(1) = 1: c(1) + 1^2/2 = 1 c + 1/2 = 1 c = 1/2. So, f(x) = x/2 + x^2/2. Then f(3) = 3/2 + 3^2/2 = 3/2 + 9/2 = 12/2 = 6.

Solution

Correct: C

Let the centers of the two circles be O1 and O2, and their radii be r1 = 5 and r2 = 3, respectively. The distance between their centers is d = O1O2 = 7. The common chord AB is perpendicular to the line segment O1O2 and is bisected by it. Let M be the midpoint of AB. So, AB = 2 * AM. Consider the triangle O1AO2. Its sides are r1=5, r2=3, and d=7. The area of this triangle can be calculated using Heron's formula. First, find the semi-perimeter s = (r1 + r2 + d) / 2 = (5 + 3 + 7) / 2 = 15/2. Area of O1AO2 = sqrt(s * (s-r1) * (s-r2) * (s-d)) Area = sqrt((15/2) * (15/2 - 5) * (15/2 - 3) * (15/2 - 7)) Area = sqrt((15/2) * (5/2) * (9/2) * (1/2)) Area = sqrt((15 * 5 * 9 * 1) / (2 * 2 * 2 * 2)) Area = sqrt(675 / 16) Area = (sqrt(225 * 3)) / 4 = (15 * sqrt(3)) / 4. We can also express the area of triangle O1AO2 in terms of its base O1O2 and height AM. The height corresponding to the base O1O2 is AM, since AM is the altitude from A to O1O2. Area of O1AO2 = 1/2 * base * height = 1/2 * O1O2 * AM = 1/2 * 7 * AM. Equating the two expressions for the area: (15 * sqrt(3)) / 4 = 1/2 * 7 * AM Multiply both sides by 2: (15 * sqrt(3)) / 2 = 7 * AM AM = (15 * sqrt(3)) / (2 * 7) = (15 * sqrt(3)) / 14. The length of the common chord AB is 2 * AM: AB = 2 * (15 * sqrt(3)) / 14 = (15 * sqrt(3)) / 7.

Solution

Correct: B

We are looking for positive integer solutions (x, y) to the equation 2x + 3y = 100. From the equation, 2x = 100 - 3y. Since 2x is an even number and 100 is an even number, 3y must also be an even number. For 3y to be even, y must be an even integer. Also, x and y must be positive integers, so x >= 1 and y >= 1. Since 2x > 0, we have 100 - 3y > 0 => 3y < 100 => y < 100/3 => y < 33.33... Since y must be an even positive integer, the possible values for y are 2, 4, 6, ..., up to the largest even integer less than 33.33, which is 32. Let y = 2k for some positive integer k. Substitute y = 2k into the equation: 2x + 3(2k) = 100 2x + 6k = 100 Divide by 2: x + 3k = 50 x = 50 - 3k. Since x must be a positive integer, x >= 1: 50 - 3k >= 1 49 >= 3k k <= 49/3 k <= 16.33... Since k must be a positive integer (because y = 2k and y >= 1 implies k >= 1/2), the possible integer values for k are 1, 2, 3, ..., 16. Each value of k corresponds to a unique pair (x, y). For example: If k=1, y=2, x=50-3(1)=47. (2*47 + 3*2 = 94 + 6 = 100) If k=16, y=32, x=50-3(16)=50-48=2. (2*2 + 3*32 = 4 + 96 = 100) There are 16 possible integer values for k, so there are 16 pairs of positive integers (x, y) that satisfy the equation.

Solution

Correct: A

We are given the sum of the first n terms of an arithmetic progression as S_n = 3n^2 - n. The nth term of an arithmetic progression, a_n, can be found using the formula a_n = S_n - S_{n-1} for n > 1. For the first term, a_1 = S_1. Let's find the first term, a_1: a_1 = S_1 = 3(1)^2 - 1 = 3 - 1 = 2. To find the 5th term, a_5, we use: a_5 = S_5 - S_4. First, calculate S_5: S_5 = 3(5)^2 - 5 = 3(25) - 5 = 75 - 5 = 70. Next, calculate S_4: S_4 = 3(4)^2 - 4 = 3(16) - 4 = 48 - 4 = 44. Now, calculate a_5: a_5 = S_5 - S_4 = 70 - 44 = 26. Alternatively, we could find the common difference d. a_2 = S_2 - S_1. S_2 = 3(2)^2 - 2 = 3(4) - 2 = 12 - 2 = 10. a_2 = 10 - 2 = 8. The common difference d = a_2 - a_1 = 8 - 2 = 6. Now, use the formula for the nth term of an arithmetic progression: a_n = a_1 + (n-1)d. a_5 = a_1 + (5-1)d = a_1 + 4d. a_5 = 2 + 4(6) = 2 + 24 = 26. Both methods yield the same result.

Solution

Correct: B

Total number of marbles in the bag = 5 red + 3 blue = 8 marbles. We are drawing two marbles at random without replacement. First, let's find the total number of ways to choose 2 marbles from 8. This is given by the combination formula C(n, k) = n! / (k! * (n-k)!). Total ways to choose 2 marbles from 8 = C(8,2) = 8! / (2! * 6!) = (8 * 7) / (2 * 1) = 28. Next, we need to find the number of ways to draw two marbles of different colors. This means one red marble AND one blue marble. Number of ways to choose 1 red marble from 5 = C(5,1) = 5. Number of ways to choose 1 blue marble from 3 = C(3,1) = 3. Number of ways to choose one red and one blue marble = C(5,1) * C(3,1) = 5 * 3 = 15. Finally, the probability is the ratio of favorable outcomes to the total possible outcomes: Probability = (Number of ways to choose different colors) / (Total number of ways to choose 2 marbles) Probability = 15 / 28.

Solution

Correct: B

Let the cone have base radius r_c = 3 and height h_c = 4. The slant height L of the cone can be found using the Pythagorean theorem: L = sqrt(r_c^2 + h_c^2) = sqrt(3^2 + 4^2) = sqrt(9 + 16) = sqrt(25) = 5. Consider a 2D cross-section of the cone and the inscribed sphere, taken through the axis of the cone. This cross-section forms an isosceles triangle with base 2*r_c = 6 and height h_c = 4. The sides of this triangle are the two slant heights (each of length L=5) and the diameter of the base (length 6). The inscribed sphere appears as an inscribed circle within this isosceles triangle. The radius of this inscribed circle is the radius of the sphere, let's call it R. The formula for the inradius R of a triangle is R = Area / s, where 's' is the semi-perimeter of the triangle. First, calculate the area of the triangular cross-section: Area = 1/2 * base * height = 1/2 * (2*r_c) * h_c = 1/2 * 6 * 4 = 12. Next, calculate the semi-perimeter 's' of the triangular cross-section: The sides of the triangle are 5, 5, and 6. s = (5 + 5 + 6) / 2 = 16 / 2 = 8. Now, calculate the radius R of the inscribed sphere: R = Area / s = 12 / 8 = 3/2. Thus, the radius of the inscribed sphere is 3/2.

Solution

Correct: E

We need to find the integer solutions for the inequality x^2 - 7x + 10 <= 0. First, factor the quadratic expression: x^2 - 7x + 10 = (x-2)(x-5). So the inequality becomes (x-2)(x-5) <= 0. The roots of the quadratic equation x^2 - 7x + 10 = 0 are x = 2 and x = 5. Since the quadratic is a parabola opening upwards (coefficient of x^2 is positive), its value is less than or equal to zero between its roots. Therefore, the inequality (x-2)(x-5) <= 0 is satisfied when 2 <= x <= 5. We are looking for integer solutions within this range. The integers are: x = 2 x = 3 x = 4 x = 5 There are 4 integer solutions.

Solution

Correct: B

The number 123 in base b can be written in base 10 as: 1 * b^2 + 2 * b^1 + 3 * b^0 = b^2 + 2b + 3. We are given that this is equal to 38 in base 10: b^2 + 2b + 3 = 38. Rearrange the equation into a standard quadratic form: b^2 + 2b + 3 - 38 = 0 b^2 + 2b - 35 = 0. We can solve this quadratic equation by factoring or using the quadratic formula. Factoring: We look for two numbers that multiply to -35 and add to 2. These numbers are 7 and -5. So, (b + 7)(b - 5) = 0. This gives two possible solutions for b: b = -7 or b = 5. Since a base 'b' must be a positive integer, b = -7 is not a valid base. Also, the digits used in base b must be less than b. In 123_b, the largest digit is 3, so b must be greater than 3. Therefore, the only valid solution is b = 5. Let's verify: In base 5, 123_5 = 1*5^2 + 2*5^1 + 3*5^0 = 1*25 + 2*5 + 3*1 = 25 + 10 + 3 = 38.

Solution

Correct: C

We are given the equation (a+b+c)(a+b-c) = 3ab. Let's expand the left side of the equation. We can group (a+b) as a single term: ( (a+b) + c ) ( (a+b) - c ) = 3ab. This is in the form (X+Y)(X-Y) = X^2 - Y^2, where X = (a+b) and Y = c. So, (a+b)^2 - c^2 = 3ab. Expand (a+b)^2: a^2 + 2ab + b^2 - c^2 = 3ab. Rearrange the terms to isolate a^2 + b^2 - c^2: a^2 + b^2 - c^2 = 3ab - 2ab a^2 + b^2 - c^2 = ab. Now, recall the Law of Cosines, which states that for a triangle with sides a, b, c and angle C opposite to side c: c^2 = a^2 + b^2 - 2ab * cos(C). We can rearrange the Law of Cosines to express a^2 + b^2 - c^2: a^2 + b^2 - c^2 = 2ab * cos(C). Now we have two expressions for a^2 + b^2 - c^2. Equate them: ab = 2ab * cos(C). Since a and b are side lengths of a triangle, they must be positive, so ab cannot be zero. We can divide both sides by ab: 1 = 2 * cos(C) cos(C) = 1/2. The angle C in a triangle must be between 0 and 180 degrees. The angle C for which cos(C) = 1/2 is 60 degrees. Therefore, the measure of angle C is 60 degrees.

Solution

Correct: C

We need to calculate (1 + i)^10. It's often easier to first calculate a smaller power and then raise it to the remaining power. Let's calculate (1 + i)^2: (1 + i)^2 = 1^2 + 2(1)(i) + i^2 Since i^2 = -1: (1 + i)^2 = 1 + 2i - 1 = 2i. Now we can rewrite (1 + i)^10 as ((1 + i)^2)^5: (1 + i)^10 = (2i)^5. Next, we apply the exponent to both the coefficient and the imaginary unit: (2i)^5 = 2^5 * i^5. Calculate 2^5: 2^5 = 32. Calculate i^5. The powers of i follow a cycle of 4: i^1 = i i^2 = -1 i^3 = -i i^4 = 1 i^5 = i^4 * i = 1 * i = i. Substitute these values back: (1 + i)^10 = 32 * i = 32i.

Solution

Correct: B

A shortest path from (0,0) to (3,3) involves only moving right (R) or up (U). To reach (3,3) from (0,0), you must make exactly 3 steps to the right and 3 steps up, for a total of 6 steps. The total number of shortest paths from (0,0) to (3,3) is given by the combination C(total steps, steps in one direction) = C(6,3). C(6,3) = 6! / (3! * 3!) = (6 * 5 * 4) / (3 * 2 * 1) = 20. Next, we need to find the number of paths that DO pass through the point (1,1) and subtract them from the total. A path passing through (1,1) can be broken into two independent parts: 1. A shortest path from (0,0) to (1,1). 2. A shortest path from (1,1) to (3,3). Number of paths from (0,0) to (1,1): This requires 1 Right step and 1 Up step, for a total of 2 steps. The number of paths is C(2,1) = 2! / (1! * 1!) = 2. Number of paths from (1,1) to (3,3): This requires (3-1) = 2 Right steps and (3-1) = 2 Up steps, for a total of 4 steps. The number of paths is C(4,2) = 4! / (2! * 2!) = (4 * 3) / (2 * 1) = 6. Total number of paths that pass through (1,1) = (paths from (0,0) to (1,1)) * (paths from (1,1) to (3,3)) = 2 * 6 = 12. Finally, the number of shortest paths from (0,0) to (3,3) that do NOT pass through (1,1) is: Total paths - Paths passing through (1,1) = 20 - 12 = 8.

Solution

Correct: B

We are given two logarithmic equations: 1) log_x (y) = 2 2) log_y (z) = 3 We want to find log_x (z). First, convert the logarithmic equations into exponential form: From equation (1): log_x (y) = 2 means y = x^2. From equation (2): log_y (z) = 3 means z = y^3. Now, substitute the expression for y from the first exponential equation into the second exponential equation: z = (x^2)^3. Using the exponent rule (a^m)^n = a^(m*n): z = x^(2*3) z = x^6. Finally, convert this back to logarithmic form with base x: log_x (z) = log_x (x^6). Using the logarithm property log_b (b^k) = k: log_x (z) = 6. Alternatively, using the change of base formula: log_x (z) = log_x (y) * log_y (z) is not a general rule. The rule is log_a (c) = log_a (b) * log_b (c). So, log_x (z) = log_x (y) * log_y (z) is correct given the base chaining. log_x (z) = 2 * 3 = 6.

Solution

Correct: A

The area of the region inside the triangle but outside all three circles can be found by subtracting the area of the circular sectors within the triangle from the total area of the equilateral triangle. 1. Area of the equilateral triangle: The formula for the area of an equilateral triangle with side length s is (sqrt(3)/4)s^2. 2. Area of the circular sectors: Each circle is centered at a vertex of the equilateral triangle, and its radius is r = s/2. The angle at each vertex of an equilateral triangle is 60 degrees. Therefore, each circular sector inside the triangle forms a 60-degree sector of a circle with radius s/2. The area of one such sector is given by (theta / 360) * pi * r^2, where theta is the angle in degrees. Area of one sector = (60 / 360) * pi * (s/2)^2 Area of one sector = (1/6) * pi * (s^2/4) Area of one sector = (pi * s^2) / 24. Since there are three such sectors (one at each vertex) and they do not overlap within the triangle (as the radius s/2 means the circles only touch at the midpoints of the sides, they don't 'cross over' into each other's sectors near the center of the triangle), The total area covered by the three sectors = 3 * (pi * s^2) / 24 = (pi * s^2) / 8. 3. Area of the desired region: Area (inside triangle, outside circles) = Area (triangle) - Total Area (three sectors) = (sqrt(3)/4)s^2 - (pi/8)s^2 = (sqrt(3)/4 - pi/8)s^2. This expression matches option A.

Solution

Correct: B

We want to find the number of positive integers n such that n^2 + 1 is divisible by n + 1. This means (n^2 + 1) / (n + 1) must be an integer. We can use polynomial long division or algebraic manipulation to simplify the expression n^2 + 1 in terms of n + 1. Consider the identity a^2 - b^2 = (a-b)(a+b). We can write n^2 + 1 as n^2 - 1 + 2. The term n^2 - 1 is divisible by n+1: n^2 - 1 = (n-1)(n+1). So, n^2 + 1 = (n-1)(n+1) + 2. For n^2 + 1 to be divisible by n + 1, it must be that (n-1)(n+1) + 2 is divisible by n + 1. Since (n-1)(n+1) is clearly divisible by n + 1, for the entire expression to be divisible by n + 1, the remainder, which is 2, must be divisible by n + 1. Therefore, n + 1 must be a divisor of 2. The divisors of 2 are {1, 2, -1, -2}. Since n is a positive integer, n must be at least 1. This implies that n + 1 must be at least 1 + 1 = 2. So, we only consider the positive divisors of 2 that are greater than or equal to 2: 1. n + 1 = 1 => n = 0. This is not a positive integer. 2. n + 1 = 2 => n = 1. This is a positive integer. Thus, there is only one positive integer n (which is n=1) for which n^2 + 1 is divisible by n + 1. Let's check for n=1: (1^2 + 1) / (1 + 1) = 2 / 2 = 1, which is an integer.

Solution

Correct: E

We are given the system of equations: 1) x + 1/y = 4 2) y + 1/x = 1 We want to find the value of xy. From equation (1), multiply by y (assuming y is not 0): xy + 1 = 4y xy = 4y - 1 (Equation 3) From equation (2), multiply by x (assuming x is not 0): xy + 1 = x xy = x - 1 (Equation 4) Now we have two expressions for xy. We can set them equal to each other: 4y - 1 = x - 1 4y = x. Substitute x = 4y into Equation 2: y + 1/(4y) = 1. To eliminate the fraction, multiply the entire equation by 4y (assuming y is not 0): 4y^2 + 1 = 4y. Rearrange this into a quadratic equation: 4y^2 - 4y + 1 = 0. This is a perfect square trinomial: (2y - 1)^2 = 0. Solving for y: 2y - 1 = 0 2y = 1 y = 1/2. Now find x using x = 4y: x = 4 * (1/2) = 2. Finally, find the value of xy: xy = 2 * (1/2) = 1. We can check these values in the original equations: 1) x + 1/y = 2 + 1/(1/2) = 2 + 2 = 4 (Correct) 2) y + 1/x = 1/2 + 1/2 = 1 (Correct) Both equations are satisfied.

Solution

Correct: A

Let N_k be the number of integers from 1 to 1000 that are divisible by k. N_k = floor(1000/k). 1. Number of integers divisible by 3: N_3 = floor(1000/3) = 333. 2. Number of integers divisible by 5: N_5 = floor(1000/5) = 200. 3. Number of integers divisible by 15 (which means divisible by both 3 and 5, since 15 is the least common multiple of 3 and 5): N_15 = floor(1000/15) = 66. We are looking for integers that are divisible by 3 or 5, but not by 15. This can be broken down using the Principle of Inclusion-Exclusion. First, find the number of integers divisible by 3 or 5 (N_3 U N_5): N_3 U N_5 = N_3 + N_5 - N_15 N_3 U N_5 = 333 + 200 - 66 N_3 U N_5 = 533 - 66 = 467. This set includes all numbers divisible by 3, all numbers divisible by 5, and counts numbers divisible by 15 only once. Now, we need to find the numbers in this set that are NOT divisible by 15. Since the set of numbers divisible by 15 (N_15) is a subset of (N_3 U N_5), we simply subtract N_15 from N_3 U N_5. Number of integers divisible by 3 or 5 but not by 15 = (N_3 U N_5) - N_15 = 467 - 66 = 401. Alternatively, consider numbers divisible by 3 only, and numbers divisible by 5 only: Numbers divisible by 3 but not 5: N_3 - N_15 = 333 - 66 = 267. Numbers divisible by 5 but not 3: N_5 - N_15 = 200 - 66 = 134. The sum of these two categories gives the desired count: 267 + 134 = 401.

Solution

Correct: A

We are given x = sqrt(3 + sqrt(5)) - sqrt(3 - sqrt(5)). We need to find x^2. Let A = sqrt(3 + sqrt(5)) and B = sqrt(3 - sqrt(5)). Then x = A - B. x^2 = (A - B)^2 = A^2 - 2AB + B^2. Let's calculate each term: A^2 = (sqrt(3 + sqrt(5)))^2 = 3 + sqrt(5). B^2 = (sqrt(3 - sqrt(5)))^2 = 3 - sqrt(5). Now, calculate 2AB: 2AB = 2 * sqrt(3 + sqrt(5)) * sqrt(3 - sqrt(5)) Using the property sqrt(a) * sqrt(b) = sqrt(a*b): 2AB = 2 * sqrt((3 + sqrt(5)) * (3 - sqrt(5))). Inside the square root, we have a product of the form (a+b)(a-b) = a^2 - b^2: (3 + sqrt(5))(3 - sqrt(5)) = 3^2 - (sqrt(5))^2 = 9 - 5 = 4. So, 2AB = 2 * sqrt(4) = 2 * 2 = 4. Now substitute A^2, B^2, and 2AB back into the expression for x^2: x^2 = (3 + sqrt(5)) + (3 - sqrt(5)) - 4. Combine like terms: x^2 = 3 + 3 + sqrt(5) - sqrt(5) - 4 x^2 = 6 - 4 x^2 = 2. Thus, x^2 is equal to 2.

Solution

Correct: B

The problem states that A, B, C, D are points on a circle in that order. This means that ABCD forms a cyclic quadrilateral. A fundamental property of cyclic quadrilaterals is that their opposite angles are supplementary (they sum to 180 degrees). In the cyclic quadrilateral ABCD: - Angle ABC is opposite to Angle ADC. - Angle BCD is opposite to Angle BAD (or DAB). We are given angle ABC = 110 degrees. According to the property of cyclic quadrilaterals: Angle ABC + Angle ADC = 180 degrees. 110 degrees + Angle ADC = 180 degrees. Angle ADC = 180 degrees - 110 degrees = 70 degrees. The information about angle BCD = 95 degrees is extra information, which can be used to find angle DAB (180 - 95 = 85 degrees) and confirm consistency if needed. (110+95+70+85 = 360).

Solution

Correct: B

Let the roots of the cubic polynomial P(x) = x^3 - 6x^2 + kx - 8 be a/r, a, and ar, since they are in geometric progression. We use Vieta's formulas to relate the roots to the coefficients of the polynomial: 1. Sum of the roots: (a/r) + a + ar = -(-6)/1 = 6. Factor out 'a': a(1/r + 1 + r) = 6. (Equation 1) 2. Sum of the products of the roots taken two at a time: (a/r)*a + a*ar + ar*(a/r) = k/1 = k. Simplify: a^2/r + a^2r + a^2 = k. (Equation 2) 3. Product of the roots: (a/r) * a * ar = -(-8)/1 = 8. Simplify: a^3 = 8. Taking the cube root, we find a = 2 (since the coefficients are real, 'a' must be real). Now substitute a = 2 into Equation 1: 2(1/r + 1 + r) = 6 Divide by 2: 1/r + 1 + r = 3 Subtract 1 from both sides: 1/r + r = 2. To solve for r, multiply the entire equation by r (assuming r is not 0, which it cannot be if the roots are non-zero): 1 + r^2 = 2r Rearrange into a quadratic equation: r^2 - 2r + 1 = 0. This is a perfect square trinomial: (r - 1)^2 = 0. Thus, r = 1. Now we have the values for a and r: a = 2 and r = 1. The roots of the polynomial are a/r = 2/1 = 2, a = 2, and ar = 2*1 = 2. So the roots are 2, 2, 2. Finally, substitute a = 2 and r = 1 into Equation 2 to find k: k = a^2/r + a^2r + a^2 k = 2^2/1 + 2^2*1 + 2^2 k = 4 + 4 + 4 k = 12. Alternatively, knowing the roots are 2, 2, 2, the polynomial is (x-2)^3. (x-2)^3 = x^3 - 3(x^2)(2) + 3(x)(2^2) - 2^3 = x^3 - 6x^2 + 12x - 8. Comparing this to P(x) = x^3 - 6x^2 + kx - 8, we see that k = 12.

Solution

Correct: B

We need to find the last digit of 7^(7^7), which is equivalent to finding 7^(7^7) modulo 10. First, let's look at the cycle of the last digits of powers of 7: 7^1 = 7 7^2 = 49 -> 9 7^3 = 343 -> 3 7^4 = 2401 -> 1 7^5 = 16807 -> 7 The cycle of the last digits is (7, 9, 3, 1), and its length is 4. To find the last digit of 7^(exponent), we need to find the exponent modulo the cycle length, which is 4. In this case, the exponent is 7^7. So we need to calculate 7^7 (mod 4). First, find 7 (mod 4): 7 is congruent to 3 (mod 4). So, 7^7 (mod 4) is congruent to 3^7 (mod 4). Now, let's look at the cycle of the last digits of powers of 3 modulo 4: 3^1 = 3 (mod 4) 3^2 = 9 is congruent to 1 (mod 4) 3^3 = 3^2 * 3 is congruent to 1 * 3 = 3 (mod 4) The cycle of last digits (mod 4) is (3, 1), and its length is 2. To find 3^7 (mod 4), we need to find the exponent 7 modulo the cycle length, which is 2. 7 is congruent to 1 (mod 2). So, 3^7 (mod 4) is congruent to 3^1 (mod 4), which is 3. This means that the exponent of 7 in the original problem (7^7) is congruent to 3 (mod 4). Therefore, the last digit of 7^(7^7) is the same as the last digit of 7^3. 7^3 = 343. The last digit of 343 is 3. Thus, the last digit of 7^7^7 is 3.

Solution

Correct: C

The graph of y = |x-2| is a V-shaped graph with its vertex at x=2, y=0. This is because |x-2| is 0 when x=2. The line y = 4 is a horizontal line. To find the area of the region bounded by these two graphs, we first need to find the points of intersection. Set y = |x-2| equal to y = 4: |x-2| = 4. This equation splits into two cases: Case 1: x - 2 = 4 => x = 6. Case 2: x - 2 = -4 => x = -2. So the two intersection points are (6,4) and (-2,4). The region bounded by y = |x-2| and y = 4 forms a triangle. The vertices of this triangle are: 1. The vertex of the absolute value function: (2,0). 2. The two intersection points: (-2,4) and (6,4). To find the area of this triangle, we can use the formula Area = 1/2 * base * height. The base of the triangle lies along the horizontal line y = 4, stretching from x = -2 to x = 6. Base length = (x_right - x_left) = 6 - (-2) = 6 + 2 = 8. The height of the triangle is the perpendicular distance from the vertex (2,0) to the line segment forming the base (which is on the line y=4). Height = (y_base - y_vertex) = 4 - 0 = 4. Now calculate the area: Area = 1/2 * base * height = 1/2 * 8 * 4 = 1/2 * 32 = 16. The area of the region is 16.