amc math - Jan 14, 07:21

Solution

Correct: C

We are given that n divides (n+1)(n+2)(n+3). We can expand the expression: (n+1)(n+2)(n+3) = (n^2 + 3n + 2)(n+3) = n^3 + 3n^2 + 2n + 3n^2 + 9n + 6 = n^3 + 6n^2 + 11n + 6. For n to divide this expression, n must divide the constant term, which is 6 (since n divides n^3, 6n^2, and 11n). So, n must be a divisor of 6. The positive divisors of 6 are 1, 2, 3, and 6. Thus, there are 4 such positive integers.

Solution

Correct: C

The recurrence relation is a_n = 2a_{n-1} + 1. We can rewrite this as a_n + 1 = 2a_{n-1} + 2, which simplifies to a_n + 1 = 2(a_{n-1} + 1). Let b_n = a_n + 1. Then the recurrence relation becomes b_n = 2b_{n-1}. This means that the sequence b_n is a geometric progression with a common ratio of 2. The first term b_1 = a_1 + 1 = 1 + 1 = 2. So, b_n = b_1 * 2^(n-1) = 2 * 2^(n-1) = 2^n. Since b_n = a_n + 1, we have a_n = b_n - 1 = 2^n - 1. We need to find the smallest n such that a_n > 1000. So, 2^n - 1 > 1000. 2^n > 1001. Let's check powers of 2: 2^1 = 2 2^2 = 4 2^3 = 8 2^4 = 16 2^5 = 32 2^6 = 64 2^7 = 128 2^8 = 256 2^9 = 512 2^10 = 1024 The smallest integer n for which 2^n > 1001 is n = 10.

Solution

Correct: C

We can use Apollonius's Theorem, which states that for a triangle ABC with median AM, AB^2 + AC^2 = 2(AM^2 + BM^2). Given AB = 7, AC = 9, and BC = 8. M is the midpoint of BC, so BM = MC = BC/2 = 8/2 = 4. Substitute the values into Apollonius's Theorem: 7^2 + 9^2 = 2(AM^2 + 4^2) 49 + 81 = 2(AM^2 + 16) 130 = 2(AM^2 + 16) Divide by 2: 65 = AM^2 + 16 Subtract 16 from both sides: AM^2 = 65 - 16 AM^2 = 49 AM = sqrt(49) AM = 7. Alternatively, using the Law of Cosines: First, find cos(B) in triangle ABC: AC^2 = AB^2 + BC^2 - 2(AB)(BC)cos(B) 9^2 = 7^2 + 8^2 - 2(7)(8)cos(B) 81 = 49 + 64 - 112cos(B) 81 = 113 - 112cos(B) 112cos(B) = 113 - 81 112cos(B) = 32 cos(B) = 32/112 = 2/7. Now, apply the Law of Cosines to triangle ABM to find AM: AM^2 = AB^2 + BM^2 - 2(AB)(BM)cos(B) AM^2 = 7^2 + 4^2 - 2(7)(4)(2/7) AM^2 = 49 + 16 - 2(4)(2) AM^2 = 65 - 16 AM^2 = 49 AM = 7.

Solution

Correct: C

There are 52 cards in a standard deck, and 4 of them are aces. Method 1: Permutations The total number of ways to choose an ordered pair of cards for the top and bottom positions is P(52, 2) = 52 * 51 = 2652. The number of ways to choose an ordered pair of aces for the top and bottom positions is P(4, 2) = 4 * 3 = 12. The probability is the ratio of favorable outcomes to the total possible outcomes: Probability = 12 / 2652 = 1 / 221. Method 2: Conditional Probability Let A be the event that the top card is an ace. Let B be the event that the bottom card is an ace. We want to find P(A and B) = P(A) * P(B|A). The probability that the top card is an ace is P(A) = 4/52 = 1/13. If the top card is an ace, then there are 51 cards remaining, and 3 of them are aces. The probability that the bottom card is an ace, given that the top card is an ace, is P(B|A) = 3/51 = 1/17. So, the probability that both the top and bottom cards are aces is P(A and B) = (1/13) * (1/17) = 1/221.

Solution

Correct: B

Let y = x^2 + 1. From this substitution, we can express x^2 in terms of y: x^2 = y - 1. Now, substitute y and x^2 into the given equation P(x^2 + 1) = x^4 + 4x^2 + 8. Since x^4 = (x^2)^2, we have x^4 = (y - 1)^2. So, P(y) = (y - 1)^2 + 4(y - 1) + 8. Expand and simplify: P(y) = (y^2 - 2y + 1) + (4y - 4) + 8 P(y) = y^2 - 2y + 1 + 4y - 4 + 8 P(y) = y^2 + (4y - 2y) + (1 - 4 + 8) P(y) = y^2 + 2y + 5. Therefore, P(x) = x^2 + 2x + 5.

Solution

Correct: C

We need to find 3^2023 (mod 7). Let's look at the powers of 3 modulo 7: 3^1 = 3 (mod 7) 3^2 = 9 = 2 (mod 7) 3^3 = 3 * 2 = 6 = -1 (mod 7) 3^4 = 3 * (-1) = -3 = 4 (mod 7) 3^5 = 3 * 4 = 12 = 5 (mod 7) 3^6 = 3 * 5 = 15 = 1 (mod 7) The powers of 3 modulo 7 repeat in a cycle of length 6 (3, 2, 6, 4, 5, 1). To find 3^2023 (mod 7), we need to find the exponent 2023 modulo 6. 2023 / 6 = 337 with a remainder of 1. So, 2023 = 6 * 337 + 1. Therefore, 3^2023 = 3^(6*337 + 1) = (3^6)^337 * 3^1 (mod 7). Since 3^6 = 1 (mod 7): 3^2023 = (1)^337 * 3^1 (mod 7) 3^2023 = 1 * 3 (mod 7) 3^2023 = 3 (mod 7). The remainder when 3^2023 is divided by 7 is 3.

Solution

Correct: D

The side length of the square ABCD is 10. Area of square ABCD = side * side = 10 * 10 = 100. Point P is on AB such that AP = 4. Since AB = 10, PB = AB - AP = 10 - 4 = 6. Point Q is on BC such that BQ = 3. Since BC = 10, QC = BC - BQ = 10 - 3 = 7. We can find the area of triangle PDQ by subtracting the areas of the three right triangles surrounding it from the area of the square: Area(PDQ) = Area(Square ABCD) - Area(Triangle ADP) - Area(Triangle BPQ) - Area(Triangle CDQ). 1. Area(Triangle ADP): This is a right triangle with legs AD and AP. AD = 10 (side of square). AP = 4 (given). Area(ADP) = (1/2) * base * height = (1/2) * AD * AP = (1/2) * 10 * 4 = 20. 2. Area(Triangle BPQ): This is a right triangle with legs PB and BQ. PB = 6 (calculated). BQ = 3 (given). Area(BPQ) = (1/2) * PB * BQ = (1/2) * 6 * 3 = 9. 3. Area(Triangle CDQ): This is a right triangle with legs CD and QC. CD = 10 (side of square). QC = 7 (calculated). Area(CDQ) = (1/2) * CD * QC = (1/2) * 10 * 7 = 35. Now, calculate Area(PDQ): Area(PDQ) = 100 - 20 - 9 - 35 Area(PDQ) = 100 - (20 + 9 + 35) Area(PDQ) = 100 - 64 Area(PDQ) = 36.

Solution

Correct: B

We need to find the integers x that satisfy both inequalities. First inequality: x^2 - 6x + 8 < 0 Factor the quadratic expression: (x - 2)(x - 4) < 0. This inequality holds when x is between the roots 2 and 4. So, 2 < x < 4. The only integer satisfying this condition is x = 3. Second inequality: |x - 3| <= 1 This absolute value inequality can be rewritten as: -1 <= x - 3 <= 1 Add 3 to all parts of the inequality: -1 + 3 <= x <= 1 + 3 2 <= x <= 4. The integers satisfying this condition are x = 2, 3, 4. Now, we need to find the integers that satisfy *both* conditions. We look for the intersection of the solution sets. Solution set for first inequality: {3} Solution set for second inequality: {2, 3, 4} Intersection: {3} intersection {2, 3, 4} = {3}. There is only 1 integer solution, which is x = 3.

Solution

Correct: B

We are given the functional equation: f(x) + 2f(1/x) = 3x (Equation 1) We want to find f(2). Let's substitute x = 2 into Equation 1: f(2) + 2f(1/2) = 3(2) f(2) + 2f(1/2) = 6 (Equation 2) Now, to get another equation involving f(2) and f(1/2), let's substitute x = 1/2 into Equation 1: f(1/2) + 2f(1/(1/2)) = 3(1/2) f(1/2) + 2f(2) = 3/2 (Equation 3) We now have a system of two linear equations with two unknowns, f(2) and f(1/2). Let A = f(2) and B = f(1/2). Equation 2 becomes: A + 2B = 6 Equation 3 becomes: B + 2A = 3/2 From Equation 3, we can express B in terms of A: B = 3/2 - 2A. Substitute this expression for B into the first equation: A + 2(3/2 - 2A) = 6 A + 3 - 4A = 6 -3A + 3 = 6 -3A = 6 - 3 -3A = 3 A = -1. Since A = f(2), we have f(2) = -1.

Solution

Correct: B

Let the angles be A, B, C. Since they are in arithmetic progression, we can write them as B-d, B, B+d for some common difference d. The sum of angles in a triangle is 180 degrees: (B-d) + B + (B+d) = 180. 3B = 180, so B = 60 degrees. Let the sides opposite angles A, B, C be a, b, c respectively. Given that the area of the triangle is 18. Area = (1/2)ac sin(B) = 18. Since B = 60 degrees, sin(B) = sin(60) = sqrt(3)/2. (1/2)ac (sqrt(3)/2) = 18 ac (sqrt(3)/4) = 18 ac = 72 / sqrt(3) = 72 * sqrt(3) / 3 = 24sqrt(3). Given that the perimeter is 18. a + b + c = 18. Now, use the Law of Cosines for side b: b^2 = a^2 + c^2 - 2ac cos(B) b^2 = a^2 + c^2 - 2ac cos(60) b^2 = a^2 + c^2 - 2ac (1/2) b^2 = a^2 + c^2 - ac. From the perimeter, we know a + c = 18 - b. Square both sides: (a+c)^2 = (18-b)^2 a^2 + c^2 + 2ac = (18-b)^2 a^2 + c^2 = (18-b)^2 - 2ac. Substitute this expression for (a^2 + c^2) into the Law of Cosines equation: b^2 = (18-b)^2 - 2ac - ac b^2 = (18-b)^2 - 3ac. We know ac = 24sqrt(3). b^2 = (18-b)^2 - 3(24sqrt(3)) b^2 = 324 - 36b + b^2 - 72sqrt(3). The b^2 terms cancel out: 0 = 324 - 36b - 72sqrt(3) 36b = 324 - 72sqrt(3) Divide by 36: b = (324 - 72sqrt(3)) / 36 b = 324/36 - 72sqrt(3)/36 b = 9 - 2sqrt(3). The length of the side opposite angle B is 9 - 2sqrt(3).

Solution

Correct: B

First, simplify the base of the expression, (1 + i) / (1 - i): To simplify, multiply the numerator and denominator by the conjugate of the denominator, which is (1 + i): (1 + i) / (1 - i) = ( (1 + i) * (1 + i) ) / ( (1 - i) * (1 + i) ) = (1 + 2i + i^2) / (1^2 - i^2) Since i^2 = -1: = (1 + 2i - 1) / (1 - (-1)) = (2i) / (1 + 1) = 2i / 2 = i. So the expression simplifies to i^2024. Now, we need to evaluate i^2024. The powers of i cycle with a period of 4: i^1 = i i^2 = -1 i^3 = -i i^4 = 1 i^5 = i, and so on. To find i^2024, we divide the exponent by 4 and look at the remainder. 2024 / 4 = 506 with a remainder of 0. When the remainder is 0, the power is equivalent to i^4, which is 1. So, i^2024 = (i^4)^506 = 1^506 = 1. The value of the expression is 1.

Solution

Correct: D

We are given the equation log_2(log_3(log_4(x))) = 1. We will solve this by repeatedly applying the definition of logarithms: if log_b(a) = c, then a = b^c. Step 1: Apply the definition for the outermost logarithm (base 2): log_3(log_4(x)) = 2^1 log_3(log_4(x)) = 2 Step 2: Apply the definition for the next logarithm (base 3): log_4(x) = 3^2 log_4(x) = 9 Step 3: Apply the definition for the innermost logarithm (base 4): x = 4^9 To simplify 4^9, we can express 4 as 2^2: x = (2^2)^9 x = 2^(2 * 9) x = 2^18. The value of x is 2^18.

Solution

Correct: B

This is a classic 'stars and bars' problem. We are looking for the number of positive integer solutions to x + y + z = 10. Since x, y, z must be at least 1, we can use a transformation to convert this to a problem of non-negative integer solutions. Let x' = x - 1, y' = y - 1, z' = z - 1. Then x', y', z' >= 0. Substitute these into the original equation: (x' + 1) + (y' + 1) + (z' + 1) = 10 x' + y' + z' + 3 = 10 x' + y' + z' = 7. Now we need to find the number of non-negative integer solutions to x' + y' + z' = 7. Using the stars and bars formula, the number of solutions is given by C(n + k - 1, k - 1), where n is the sum (7 stars) and k is the number of variables (3 bars). Number of solutions = C(7 + 3 - 1, 3 - 1) = C(9, 2). C(9, 2) = 9! / (2! * (9-2)!) = 9! / (2! * 7!) = (9 * 8) / (2 * 1) = 72 / 2 = 36. There are 36 distinct positive integer solutions.

Solution

Correct: C

Let the centers of the two circles be O1 and O2, and their radii be r1 = 5 and r2 = 3, respectively. Since the circles are tangent externally, the distance between their centers is O1O2 = r1 + r2 = 5 + 3 = 8. Let the common external tangent touch the first circle (center O1, radius r1) at point A and the second circle (center O2, radius r2) at point B. The radii O1A and O2B are perpendicular to the tangent line AB. Draw a line through O2 parallel to AB, intersecting O1A at point C. This forms a rectangle O2BCA. Thus, AB = O2C and BC = O2A (no, BC is not O2A). Instead, consider the right trapezoid O1ABO2 (where O1A and O2B are parallel). Draw a perpendicular from O2 to O1A. Let this point be C. Then O1C = O1A - CA = O1A - O2B = r1 - r2 = 5 - 3 = 2. And O2C is parallel to AB, so O2C = AB. Now, consider the right-angled triangle O1CO2. The hypotenuse is O1O2 = 8. The leg O1C = r1 - r2 = 2. The other leg is O2C, which is equal to the length AB we want to find. By the Pythagorean theorem: O1O2^2 = O1C^2 + O2C^2 8^2 = 2^2 + AB^2 64 = 4 + AB^2 AB^2 = 64 - 4 AB^2 = 60 AB = sqrt(60) AB = sqrt(4 * 15) AB = 2sqrt(15).

Solution

Correct: A

For a number to be a perfect square, all the exponents in its prime factorization must be even. First, we need to find the prime factorization of 2023. We can test small prime numbers: 2023 is not divisible by 2, 3 (sum of digits 2+0+2+3=7), 5. Let's try 7: 2023 / 7 = 289. So, 2023 = 7 * 289. Now, we need to factorize 289. We might recognize 289 as 17^2. Indeed, 17 * 17 = 289. So, the prime factorization of 2023 is 7^1 * 17^2. We want 2023n to be a perfect square. Let's substitute the prime factorization: 2023n = (7^1 * 17^2) * n. For 2023n to be a perfect square, all the prime exponents in its factorization must be even. In 7^1 * 17^2: - The exponent of 17 is 2, which is already even. - The exponent of 7 is 1, which is odd. To make the exponent of 7 even, the smallest positive integer n must have at least one factor of 7. If n contains any other prime factors, their exponents must also become even in the product 2023n. To make n the smallest, we only include the necessary prime factors with the minimum required exponents. So, n must contain a factor of 7. If n = 7, then 2023n = 7^1 * 17^2 * 7 = 7^2 * 17^2 = (7 * 17)^2. This is a perfect square (119^2). Any other prime factor in n would make n larger. For example, if n = 7 * k^2 for some integer k > 1, then n would be larger. Therefore, the smallest positive integer n is 7.

Solution

Correct: B

We are given: 1) x + y = 5 2) x^2 + y^2 = 13 We need to find x^3 + y^3. Step 1: Find the value of xy. We know that (x + y)^2 = x^2 + 2xy + y^2. Substitute the given values into this identity: 5^2 = 13 + 2xy 25 = 13 + 2xy 2xy = 25 - 13 2xy = 12 xy = 6. Step 2: Use the sum of cubes formula. The sum of cubes formula is x^3 + y^3 = (x + y)(x^2 - xy + y^2). Substitute the known values into this formula: x^3 + y^3 = (5)(13 - 6) x^3 + y^3 = (5)(7) x^3 + y^3 = 35. Therefore, x^3 + y^3 = 35.

Solution

Correct: B

First, find the equation of the line passing through the points (1, 2) and (3, 8). Step 1: Calculate the slope (m) of the line. m = (y2 - y1) / (x2 - x1) m = (8 - 2) / (3 - 1) m = 6 / 2 m = 3. Step 2: Use the point-slope form (y - y1 = m(x - x1)) to find the equation of the line. Using point (1, 2) and slope m = 3: y - 2 = 3(x - 1) y - 2 = 3x - 3 y = 3x - 1. Step 3: The third point (x, y) is collinear with the first two points, so it must lie on the line y = 3x - 1. We are also given that this third point (x, y) lies on the line y = 2x - 1. Step 4: Find the x-coordinate of the third point by setting the two y-equations equal to each other. 3x - 1 = 2x - 1 Add 1 to both sides: 3x = 2x Subtract 2x from both sides: x = 0. To verify, if x=0, then y = 2(0) - 1 = -1. So the third point is (0, -1). Let's check if it lies on y = 3x - 1: y = 3(0) - 1 = -1. It does. Thus, the value of x is 0.

Solution

Correct: D

This is a stars and bars problem with a lower bound for each variable. We have the equation x + y + z + w = 15, with the conditions x >= 2, y >= 2, z >= 2, w >= 2. To use the standard stars and bars formula (which applies to non-negative integers), we make a substitution. Let x' = x - 2, y' = y - 2, z' = z - 2, w' = w - 2. Now, x', y', z', w' must be non-negative integers (x', y', z', w' >= 0). Substitute these into the original equation: (x' + 2) + (y' + 2) + (z' + 2) + (w' + 2) = 15 x' + y' + z' + w' + 8 = 15 x' + y' + z' + w' = 15 - 8 x' + y' + z' + w' = 7. Now we need to find the number of non-negative integer solutions to x' + y' + z' + w' = 7. Using the stars and bars formula, for an equation of the form a1 + a2 + ... + ak = n (where ai >= 0), the number of solutions is C(n + k - 1, k - 1) or C(n + k - 1, n). Here, n = 7 (the sum) and k = 4 (the number of variables). Number of solutions = C(7 + 4 - 1, 4 - 1) = C(10, 3) C(10, 3) = 10! / (3! * (10 - 3)!) = 10! / (3! * 7!) = (10 * 9 * 8) / (3 * 2 * 1) = (10 * 3 * 4) (after canceling 9/3 and 8/2) = 120. There are 120 distinct positive integer solutions.

Solution

Correct: C

1. **Analyze triangle ABC:** Given that triangle ABC is isosceles with AB = AC, and angle A = 40 degrees. The base angles are equal: angle B = angle C. Sum of angles in a triangle is 180 degrees: Angle B + Angle C + Angle A = 180 degrees 2 * Angle C + 40 degrees = 180 degrees 2 * Angle C = 140 degrees Angle C = 70 degrees. So, Angle ABC = Angle C = 70 degrees. 2. **Analyze triangle BCD:** Point D is on AC such that BD = BC. This means triangle BCD is an isosceles triangle. The base angles are equal: angle BDC = angle BCD. We know angle BCD is the same as angle C of the larger triangle, so angle BCD = 70 degrees. Therefore, angle BDC = 70 degrees. Now, find angle CBD in triangle BCD: Angle CBD + Angle BCD + Angle BDC = 180 degrees Angle CBD + 70 degrees + 70 degrees = 180 degrees Angle CBD + 140 degrees = 180 degrees Angle CBD = 40 degrees. 3. **Find angle ABD:** We want to find angle ABD. We know that angle ABC is composed of angle ABD and angle CBD. Angle ABC = Angle ABD + Angle CBD. We found Angle ABC = 70 degrees and Angle CBD = 40 degrees. 70 degrees = Angle ABD + 40 degrees Angle ABD = 70 degrees - 40 degrees Angle ABD = 30 degrees.

Solution

Correct: B

The sum is a telescoping series. Let's write out the first few terms and the last few terms of the sum: For k=1: (1/1 - 1/2) For k=2: (1/2 - 1/3) For k=3: (1/3 - 1/4) ... For k=99: (1/99 - 1/100) For k=100: (1/100 - 1/101) Now, let's sum these terms: Sum = (1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + ... + (1/99 - 1/100) + (1/100 - 1/101) Notice that the middle terms cancel each other out: -1/2 cancels with +1/2 -1/3 cancels with +1/3 ... and so on. -1/100 cancels with +1/100 The only terms that remain are the very first term and the very last term: Sum = 1/1 - 1/101 Sum = 1 - 1/101 To combine these, find a common denominator: Sum = 101/101 - 1/101 Sum = 100/101. The value of the sum is 100/101.

Solution

Correct: B

Let the prime factorization of n be p1^a1 * p2^a2 * ... * pk^ak. The number of divisors, d(n), is given by the product of (ai + 1) for each prime factor: d(n) = (a1+1)(a2+1)...(ak+1). We are looking for d(n) = 10. There are two ways to express 10 as a product of integers greater than 1: Case 1: 10 = 10 (meaning only one prime factor) In this case, a1 + 1 = 10, so a1 = 9. The number n would be of the form p^9, where p is a prime number. To make n the smallest, we choose the smallest prime number, p=2. So, n = 2^9 = 512. Case 2: 10 = 5 * 2 (meaning two distinct prime factors) In this case, we have two exponents a1 and a2 such that (a1+1)(a2+1) = 10. So, a1+1 = 5 and a2+1 = 2 (or vice versa). This means a1 = 4 and a2 = 1. The number n would be of the form p1^4 * p2^1, where p1 and p2 are distinct prime numbers. To make n the smallest, we should assign the smallest prime (2) to the higher exponent (4) and the next smallest prime (3) to the lower exponent (1). So, let p1 = 2 and p2 = 3. n = 2^4 * 3^1 = 16 * 3 = 48. If we assigned p1 = 3 and p2 = 2, then n = 3^4 * 2^1 = 81 * 2 = 162. Comparing the values of n from both cases: From Case 1: n = 512. From Case 2: n = 48 (or 162). The smallest integer n > 1 from these possibilities is 48. Let's check d(48): 48 = 16 * 3 = 2^4 * 3^1. d(48) = (4+1)(1+1) = 5 * 2 = 10. This is correct.

Solution

Correct: B

We are given two equations: 1) 2^x = 3 2) 3^y = 5 We need to find the value of 2^(xy). From equation (1), we know that 3 is equivalent to 2^x. Substitute this expression for 3 into equation (2): (2^x)^y = 5 Using the exponent rule (a^m)^n = a^(m*n): 2^(x*y) = 5 Thus, the value of 2^(xy) is 5.

Solution

Correct: D

Given that DE is parallel to BC (DE || BC). This implies that triangle ADE is similar to triangle ABC (by AA similarity, as angle A is common and angle ADE = angle ABC, angle AED = angle ACB corresponding angles). We are given AD/DB = 2/3. This means that AD = 2k and DB = 3k for some constant k. Therefore, the full side AB = AD + DB = 2k + 3k = 5k. The ratio of corresponding sides of similar triangles ADE and ABC is AD/AB. AD/AB = (2k)/(5k) = 2/5. For similar triangles, the ratio of their areas is the square of the ratio of their corresponding sides. Area(ADE) / Area(ABC) = (AD/AB)^2 Area(ADE) / Area(ABC) = (2/5)^2 Area(ADE) / Area(ABC) = 4/25. We are given that Area(ADE) = 12. So, 12 / Area(ABC) = 4/25. To find Area(ABC), cross-multiply: Area(ABC) = 12 * (25/4) Area(ABC) = (12/4) * 25 Area(ABC) = 3 * 25 Area(ABC) = 75. The trapezoid DBCE is the region remaining when triangle ADE is removed from triangle ABC. Area(Trapezoid DBCE) = Area(ABC) - Area(ADE). Area(Trapezoid DBCE) = 75 - 12 Area(Trapezoid DBCE) = 63.

Solution

Correct: C

Let the roots of the cubic polynomial P(x) = x^3 - 6x^2 + Ax - 6 be r-d, r, and r+d, since they are in an arithmetic progression. From Vieta's formulas, for a cubic polynomial ax^3 + bx^2 + cx + d = 0, we have: 1. Sum of the roots: (r-d) + r + (r+d) = -b/a 2. Sum of the products of the roots taken two at a time: (r-d)r + r(r+d) + (r-d)(r+d) = c/a 3. Product of the roots: (r-d)r(r+d) = -d/a For P(x) = x^3 - 6x^2 + Ax - 6, we have a=1, b=-6, c=A, d=-6. Using the sum of the roots: (r-d) + r + (r+d) = -(-6)/1 3r = 6 r = 2. So, one of the roots is 2. Since 2 is a root of P(x), P(2) must be equal to 0. P(2) = (2)^3 - 6(2)^2 + A(2) - 6 = 0 8 - 6(4) + 2A - 6 = 0 8 - 24 + 2A - 6 = 0 -16 + 2A - 6 = 0 2A - 22 = 0 2A = 22 A = 11. (Optional: Verify with the other Vieta's formulas to ensure consistency) If r=2 and A=11, the roots are 2-d, 2, 2+d. Product of the roots: (2-d)(2)(2+d) = -(-6)/1 = 6 2(4 - d^2) = 6 4 - d^2 = 3 d^2 = 1, so d = +/- 1. Let's choose d=1. The roots are 1, 2, 3. Sum of products of roots taken two at a time: (1)(2) + (1)(3) + (2)(3) = A/1 2 + 3 + 6 = A 11 = A. This confirms the value A=11.

Solution

Correct: C

To make N the smallest possible positive integer, we want it to have: 1. The fewest possible number of digits. 2. The leftmost (most significant) digit to be as small as possible. To achieve the fewest possible digits, we should maximize the value of each digit. Since all digits must be greater than 0, the maximum digit value is 9. Let k be the number of digits in N. The maximum sum of digits for k digits is 9k. We need the sum of digits to be 2023, so 9k >= 2023. k >= 2023/9. 2023 / 9 = 224 with a remainder of 7 (2023 = 9 * 224 + 7). So, k must be at least 224.77..., which means k must be at least 225. Therefore, N has 225 digits. Now, to make N the smallest, we want its first digit to be as small as possible. The remaining digits should be as large as possible (9s) to absorb the largest possible sum, thereby leaving the smallest possible value for the first digit. Let the first digit be d1. The sum of the remaining (k-1) = 224 digits is 2023 - d1. These 224 digits must all be between 1 and 9. To make d1 smallest, the remaining 224 digits should be as large as possible, i.e., 9s. So, the maximum possible sum for the 224 digits is 224 * 9 = 2016. Therefore, 2023 - d1 must be less than or equal to 2016 (because we can't make the sum of the remaining digits greater than their maximum possible value). 2023 - d1 <= 2016 2023 - 2016 <= d1 7 <= d1. Since all digits must be greater than 0, the possible values for d1 are 7, 8, 9. To make N the smallest integer, we choose the smallest possible first digit, which is d1 = 7. If the first digit is 7, the sum of the remaining 224 digits must be 2023 - 7 = 2016. Since the maximum sum for 224 digits is 224 * 9 = 2016, this means all the remaining 224 digits must indeed be 9s. So, the number N is 7 followed by 224 nines (799...99). This is the smallest such integer. The first digit of N is 7.