High-school contest problems: algebra, geometry, counting, and number theory at real AMC 10 difficulty. Try each problem first, then open the strategy analysis and worked solution.
Let r and s be the roots of x² − 5x + 3 = 0. What is the value of r³ + s³?
A. 65 · B. 70 · C. 80 · D. 95 · E. 110
Never solve for the roots — use Vieta's formulas plus the identity r³ + s³ = (r + s)³ − 3rs(r + s) to get the answer from the sum and product alone.
By Vieta's formulas, r + s = 5 and rs = 3. Then r³ + s³ = (r + s)³ − 3rs(r + s) = 5³ − 3(3)(5) = 125 − 45 = 80.
Answer: C
Two circles with radii 3 and 5 are externally tangent. A line is tangent to both circles, touching each circle at exactly one point. What is the distance between the two points of tangency?
A. 2√14 · B. 2√15 · C. 8 · D. 4√5 · E. 2√17
Drop a perpendicular from the smaller circle's center to the radius of the larger circle: the tangent length becomes the leg of a right triangle whose hypotenuse is the distance between centers.
Let the centers be O₁ and O₂. Since the circles are externally tangent, O₁O₂ = 3 + 5 = 8. Draw the segment through O₁ parallel to the common tangent line, meeting the radius of the larger circle at a right angle. This forms a right triangle with hypotenuse 8 and one leg equal to the difference of the radii, 5 − 3 = 2. The other leg, which equals the distance between the points of tangency, is √(8² − 2²) = √60 = 2√15.
Answer: B
How many four-digit positive integers contain exactly two digits equal to 7?
A. 432 · B. 459 · C. 486 · D. 513 · E. 540
Split on whether the leading digit is a 7 or not — the leading digit can't be zero, so the two cases have different counts for the remaining digits.
Case 1: the thousands digit is 7. Choose the position of the other 7 among the last three digits: 3 ways. The remaining two digits can each be any of the 9 digits other than 7: 9² = 81 ways. This gives 3 × 81 = 243 numbers. Case 2: the thousands digit is not 7. Choose the two positions of the 7s among the last three digits: C(3,2) = 3 ways. The thousands digit has 8 choices (1–9 except 7) and the remaining digit has 9 choices: 3 × 8 × 9 = 216 numbers. The total is 243 + 216 = 459.
Answer: B
What is the units digit of 7^(7^7)?
A. 1 · B. 7 · C. 9 · D. 3 · E. 5
Two cycles at once: the units digits of powers of 7 cycle with period 4, and the exponent 7^7 modulo 4 is found by noting 7 ≡ −1 (mod 4) raised to an odd power.
The units digits of 7¹, 7², 7³, 7⁴, … are 7, 9, 3, 1, repeating with period 4. So the units digit of 7^(7^7) depends on 7^7 mod 4. Since 7 ≡ −1 (mod 4) and 7 is odd, 7^7 ≡ (−1)^7 = −1 ≡ 3 (mod 4). The exponent is 3 mod 4, matching 7³, whose units digit is 3.
Answer: D
The sum of the first 10 terms of an arithmetic sequence is 100, and the sum of the first 20 terms is 400. What is the 15th term of the sequence?
A. 27 · B. 29 · C. 31 · D. 33 · E. 35
Subtract the two sum formulas: the difference kills the first term's contribution pattern and isolates the common difference immediately.
Let the first term be a and the common difference be d. Then S₁₀ = 5(2a + 9d) = 100, so 2a + 9d = 20. Also S₂₀ = 10(2a + 19d) = 400, so 2a + 19d = 40. Subtracting gives 10d = 20, so d = 2, and then 2a = 20 − 18 = 2, so a = 1. The 15th term is a + 14d = 1 + 28 = 29.
Answer: B
How many integers n with 1 ≤ n ≤ 100 satisfy the condition that ⌊√n⌋ divides n?
A. 24 · B. 26 · C. 28 · D. 30 · E. 32
Split the range into blocks where ⌊√n⌋ is constant: in each block [m², (m+1)² − 1] the condition becomes "m divides n," a simple multiples count.
For each m, the integers with ⌊√n⌋ = m are n = m², m² + 1, …, m² + 2m. The multiples of m in this block are m², m² + m, and m² + 2m — exactly 3 numbers. This holds for m = 1 through 9, giving 9 × 3 = 27 numbers. For m = 10, the block within range is just n = 100, and 10 divides 100, adding 1 more. The total is 27 + 1 = 28.
Answer: C
Compute 100² − 99² + 98² − 97² + ⋯ + 2² − 1².
A. 4950 · B. 5050 · C. 5100 · D. 5151 · E. 5000
Pair adjacent terms and use the difference of squares: each pair collapses to a single odd number, turning 100 squares into a short arithmetic series.
Pair the terms as (100² − 99²) + (98² − 97²) + ⋯ + (2² − 1²). Each pair is a difference of squares: k² − (k−1)² = 2k − 1. So the sum is 199 + 197 + ⋯ + 3, an arithmetic series of 50 odd numbers from 3 to 199. Its sum is 50 × (3 + 199)/2 = 50 × 101 = 5050.
Answer: B
In triangle ABC, AB = 13, BC = 14, and CA = 15. What is the length of the altitude from A to BC?
A. 10 · B. 12 · C. 13 · D. 60/7 · E. 14
Recognize the 13-14-15 triangle: Heron's formula gives a clean area of 84, and the altitude follows from area = (1/2) × base × height.
The semiperimeter is s = (13 + 14 + 15)/2 = 21. By Heron's formula, the area is √(21 × 8 × 7 × 6) = √7056 = 84. The altitude h from A to the base BC = 14 satisfies (1/2)(14)(h) = 84, so h = 12.
Answer: B
What is the smallest positive integer that has exactly 12 positive divisors?
A. 36 · B. 48 · C. 60 · D. 72 · E. 96
Work backwards from the divisor-count formula: factor 12 as 12, 6 × 2, 4 × 3, or 3 × 2 × 2, build the smallest number for each shape (largest exponents on the smallest primes), and take the minimum.
If n = p₁^e₁ ⋯ pₖ^eₖ, its divisor count is (e₁+1)⋯(eₖ+1) = 12. The possible exponent patterns, with the largest exponents assigned to the smallest primes, give: p¹¹ = 2048; p⁵q = 2⁵ × 3 = 96; p³q² = 2³ × 3² = 72; p²qr = 2² × 3 × 5 = 60. The smallest is 60, and indeed 60 = 2² × 3 × 5 has (2+1)(1+1)(1+1) = 12 divisors.
Answer: C
Real numbers a and b satisfy a + b = 4 and a² + b² = 10. What is the value of a⁴ + b⁴?
A. 78 · B. 82 · C. 86 · D. 90 · E. 100
Climb the power ladder with symmetric identities: get ab from the first two equations, then square the sum of squares and subtract twice a²b².
From (a + b)² = a² + 2ab + b², we get 16 = 10 + 2ab, so ab = 3. Then a⁴ + b⁴ = (a² + b²)² − 2a²b² = 10² − 2(3²) = 100 − 18 = 82.
Answer: B
A committee of 4 people is to be chosen from 6 men and 5 women so that at least 2 of the committee members are women. In how many ways can this be done?
A. 200 · B. 215 · C. 230 · D. 245 · E. 260
Split "at least 2 women" into exact cases (2, 3, or 4 women) and add — casework on the exact count is cleaner than subtracting the complement here.
Exactly 2 women and 2 men: C(5,2) × C(6,2) = 10 × 15 = 150. Exactly 3 women and 1 man: C(5,3) × C(6,1) = 10 × 6 = 60. Exactly 4 women: C(5,4) = 5. The total is 150 + 60 + 5 = 215.
Answer: B
What is the smallest positive integer n such that n! is divisible by 1000?
A. 10 · B. 15 · C. 20 · D. 25 · E. 30
The bottleneck is the scarcer prime: 1000 = 2³ × 5³, and factors of 5 are far rarer in a factorial than factors of 2, so count the 5s (Legendre's idea) and check the 2s take care of themselves.
Since 1000 = 2³ × 5³, n! must contain at least three factors of 5. The multiples of 5 up to n contribute ⌊n/5⌋ factors of 5 (multiples of 25 would add more, but none appear before 25). We need ⌊n/5⌋ ≥ 3, so n ≥ 15. At n = 15, the factors 5, 10, and 15 supply three 5s, and factors of 2 are plentiful, so 15! is divisible by 1000, while no smaller n works.
Answer: B
Point P lies 13 units from the center O of a circle of radius 5. A secant line through P meets the circle at points A and B, with A between P and B, and PA = 4. What is PB?
A. 32 · B. 36 · C. 40 · D. 44 · E. 48
Power of a point does the whole job in one line: PA × PB equals the power OP² − r², no need to find where the secant actually hits the circle.
The power of P with respect to the circle is OP² − r² = 13² − 5² = 169 − 25 = 144. For the secant through P, power also equals PA × PB. With PA = 4, we get 4 × PB = 144, so PB = 36.
Answer: B
For how many integers n is n² − 10n + 24 a perfect square?
A. 1 · B. 2 · C. 3 · D. 4 · E. 5
Complete the square, then move the square to the other side: a difference of two squares equal to 1 factors into two integers whose product is 1, leaving only a couple of possibilities.
Complete the square: n² − 10n + 24 = (n − 5)² − 1. Set this equal to k² for some integer k ≥ 0. Then (n − 5)² − k² = 1, so (n − 5 − k)(n − 5 + k) = 1. Both factors are integers with product 1, so they are both 1 or both −1. Both 1 gives n − 5 = 1, so n = 6. Both −1 gives n − 5 = −1, so n = 4. No other possibilities exist, so there are 2 such integers.
Answer: B