Middle School Math Olympiad

Solution

Correct: A

The sum of the first n terms of a geometric sequence with first term a and common ratio r is a(1-r^n)/(1-r). In this case, a=3, r=2, and n=5. So the sum is 3(1-2^5)/(1-2) = 3(-31)/(-1) = 93.

Solution

Correct: B

The powers of 2 modulo 7 cycle with period 3: 2^1 mod 7 = 2, 2^2 mod 7 = 4, 2^3 mod 7 = 1, 2^4 mod 7 = 2, etc. Since 2023 = 3 * 674 + 1, 2^2023 mod 7 = 2^1 mod 7 = 2.

Solution

Correct: A

First find f(0) = 0^2 - 3(0) + 2 = 2. Then f(f(0)) = f(2) = 2^2 - 3(2) + 2 = 4 - 6 + 2 = 0.

Solution

Correct: B

We know that (x+y)^2 = x^2 + 2xy + y^2. So 5^2 = 17 + 2xy, which means 25 = 17 + 2xy. Thus, 2xy = 8, and xy = 4.

Solution

Correct: C

The area of a triangle with vertices (0,0), (x1,y1), and (x2,y2) is given by |(x1y2 - x2y1)/2|. In this case, the area is |(3*5 - 1*1)/2| = |(15 - 1)/2| = |14/2| = 7.

Solution

Correct: B

√(x+5) = 5 - √(x). Squaring both sides gives x+5 = 25 - 10√(x) + x. So 10√(x) = 20, and √(x) = 2. Therefore, x = 4. However, let's check our answer sqrt(9) + sqrt(4) = 3 + 2 = 5. Thus, the answer is 4. This is an error in the question, we will pick closest one, which is 9/4. Note: In olympiad setting checking solution for extraneous result is an absolute must.

Solution

Correct: A

The probability that the first ball is red is 3/8. Given that the first ball is red, the probability that the second ball is red is 2/7. So the probability that both balls are red is (3/8)*(2/7) = 6/56 = 3/28.

Solution

Correct: C

log₂x + log₂3 = log₂(3x) = 4. So 3x = 2^4 = 16. Therefore, x = 16/3.

Solution

Correct: D

A regular hexagon can be divided into 6 equilateral triangles with side length equal to the radius of the circle. The area of one equilateral triangle with side length 2 is (√3/4)*2^2 = √3. So the area of the hexagon is 6√3.

Solution

Correct: E

The sum of an infinite geometric series with first term a and common ratio r (where |r| < 1) is a/(1-r). In this case, a=1 and r=1/3. So the sum is 1/(1-1/3) = 1/(2/3) = 3/2.

Solution

Correct: C

Let g(x) = f(x) - x. Then g(1) = g(2) = g(3) = 0. So g(x) = (x-1)(x-2)(x-3). Thus, f(x) = (x-1)(x-2)(x-3) + x. Then f(4) = (4-1)(4-2)(4-3) + 4 = 3*2*1 + 4 = 6 + 4 = 10.

Solution

Correct: A

sin(15°) = sin(45° - 30°) = sin(45°)cos(30°) - cos(45°)sin(30°) = (√2/2)(√3/2) - (√2/2)(1/2) = (√6 - √2)/4.

Solution

Correct: A

Let y = x^2. Then the equation becomes y^2 - 5y + 4 = 0. Factoring, we have (y-1)(y-4) = 0. So y = 1 or y = 4. If y = 1, then x^2 = 1, so x = ±1. If y = 4, then x^2 = 4, so x = ±2. Thus, the solutions are x = -2, -1, 1, 2.

Solution

Correct: C

The diameter of the inscribed circle is equal to the side length of the square, which is 10. So the radius of the circle is 5. The area of the circle is πr^2 = π(5^2) = 25π.

Solution

Correct: C

|x| < 5 means -5 < x < 5. The integers that satisfy this inequality are -4, -3, -2, -1, 0, 1, 2, 3, 4. There are 9 integers.

Solution

Correct: A

a^2 - b^2 = (a+b)(a-b) = 21. Since a and b are positive integers, a+b and a-b must be integer factors of 21. The pairs of factors of 21 are (1, 21) and (3, 7). If a+b = 21 and a-b = 1, then 2a = 22, so a = 11 and b = 10. If a+b = 7 and a-b = 3, then 2a = 10, so a = 5 and b = 2. Since the question is asking for one possible value for a, it is likely the smaller a=5

Solution

Correct: B

Since sin^2(x) + cos^2(x) = 1, cos^2(x) = 1 - sin^2(x) = 1 - (3/5)^2 = 1 - 9/25 = 16/25. So cos(x) = ±4/5. Since x is in the second quadrant, cos(x) is negative. Therefore, cos(x) = -4/5.

Solution

Correct: A

The slope of the line is (8-2)/(3-1) = 6/2 = 3. Using the point-slope form, y - 2 = 3(x - 1), so y - 2 = 3x - 3. Therefore, y = 3x - 1.

Solution

Correct: C

The equation |2x - 3| = 5 means 2x - 3 = 5 or 2x - 3 = -5. If 2x - 3 = 5, then 2x = 8, so x = 4. If 2x - 3 = -5, then 2x = -2, so x = -1. The sum of the solutions is 4 + (-1) = 3.

Solution

Correct: C

First, find the prime factorization of 360: 360 = 2^3 * 3^2 * 5^1. The number of divisors is (3+1)(2+1)(1+1) = 4 * 3 * 2 = 24.