AMC 8 Problem-Solving Strategies

📚 Master Guide 🎯 Target: AMC 8 Students & Coaches 📅 Updated: 2025

📑 Table of Contents

  1. Arithmetic & Number Sense
  2. Pre-Algebra & Word Problems
  3. Algebra & Logic
  4. Geometry
  5. Counting & Probability

1 Arithmetic & Number Sense

Summation & Telescoping Series

Key Ideas & When to Use

  • Trigger: Long sums of fractions (e.g., \(\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \dots\)) or products that look like they might cancel out.
  • Core Method: "Telescoping" means rewriting terms so the middle parts cancel, leaving only the first and last terms.

Core Formulas

1. Gauss Sum: \(1 + 2 + \dots + n = \frac{n(n+1)}{2}\)

2. Sum of Squares: \(1^2 + 2^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}{6}\)

3. Sum of Cubes: \(1^3 + 2^3 + \dots + n^3 = \left[\frac{n(n+1)}{2}\right]^2\)

4. Telescoping Fraction: \(\frac{1}{n(n+k)} = \frac{1}{k} \left( \frac{1}{n} - \frac{1}{n+k} \right)\)

Representative Example

Problem: Evaluate the sum \(\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \dots + \frac{1}{99\cdot100}\).

Solution:
  1. Use the identity \(\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}\).
  2. Expand: \(\left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \dots + \left(\frac{1}{99} - \frac{1}{100}\right)\).
  3. Cancel middle terms: \(1 - \frac{1}{100} = \frac{99}{100}\).

Factors, Multiples, and Primes

Key Ideas & When to Use

  • Trigger: "Number of divisors," "GCD/LCM," "Factorials."
  • Core Method: Prime Factorization is the "DNA" of the number.
  • Euclidean Algorithm: To find \(GCD(a,b)\), compute \(GCD(a-b, b)\). Repeat until one number is 0.

Mini Toolkit

  • Number of Factors: If \(N = p_1^{e_1} \dots p_k^{e_k}\), count = \((e_1+1)\dots(e_k+1)\).
  • Product of Factors: \(N^{\text{total factors} / 2}\).
  • Double Factorial (\(n!!\)): Product of same-parity integers up to \(n\).
    • \(5!! = 5 \times 3 \times 1 = 15\)
    • \(6!! = 6 \times 4 \times 2 = 48\)
Representative Example

Problem: 2020 AMC 8 #17
How many factors of 2020 have more than 3 factors?

Solution:
  1. \(2020 = 20 \times 101 = 2^2 \times 5 \times 101\).
  2. Total factors = \((2+1)(1+1)(1+1) = 12\).
  3. Numbers with \(\le 3\) factors are: 1 (1 factor), Primes (2 factors), Squares of Primes (3 factors).
  4. Primes dividing 2020: 2, 5, 101 (3 numbers).
  5. Squares of primes dividing 2020: \(2^2=4\) (1 number). \(5^2\) and \(101^2\) are too big.
  6. Identity: 1 (1 number).
  7. Excluded: \(1 + 3 + 1 = 5\).
  8. Target: \(12 - 5 = 7\). Answer: (B) 7

Modular Arithmetic & Last Digits

Key Ideas & When to Use

  • Trigger: "Remainder when divided by," "Units digit," "Day of the week."
  • Core Method: Use Cycles. Powers repeat.
  • Chinese Remainder Theorem (Logic): If \(x \equiv a \pmod m\) and \(x \equiv b \pmod n\), look for the first number in the list of one that satisfies the other.

Mini Toolkit (Last Digit Cycles)

  • Ends in 0, 1, 5, 6: Last digit stays the same.
  • Ends in 4, 9: Cycle length 2 (\(4, 6\) and \(9, 1\)).
  • Ends in 2, 3, 7, 8: Cycle length 4.
    • 2: 2, 4, 8, 6
    • 3: 3, 9, 7, 1
    • 7: 7, 9, 3, 1
    • 8: 8, 4, 2, 6
Representative Example

Problem: 2022 AMC 8 #17
Find the units digit of \(2!! + 4!! + 6!! + \dots + 2022!!\).

Solution:
  1. Note that for \(n \ge 10\), \(n!!\) contains factors 2 and 5 (from 10), so units digit is 0.
  2. We only calculate up to \(8!!\).
  3. \(2!! = 2\).
  4. \(4!! = 4 \times 2 = 8\).
  5. \(6!! = 6 \times 4 \times 2 = 48 \to 8\).
  6. \(8!! = 8 \times 6 \times 4 \times 2 = 384 \to 4\).
  7. Sum: \(2 + 8 + 8 + 4 = 22\). Units digit is 2.

2 Pre-Algebra & Word Problems

Ratios, Mixtures, and Rates

Key Ideas & When to Use

  • Trigger: "Mixture of milk and water," "Round trip speed," "Work together."
  • Trap: Do NOT average speeds (\(\frac{v_1+v_2}{2}\)). Use \(\frac{\text{Total Distance}}{\text{Total Time}}\).
  • Work Formula: \(\frac{1}{A} + \frac{1}{B} = \frac{1}{T_{\text{together}}}\).
Representative Example

Problem: 2013 AMC 8 #11 (Mixture/Rate)
Milk/Cream ratio 7:2. Total 450mL. Change to 5:3 by replacing milk with cream.

Solution:
  1. Original: 9 parts = 450mL \(\to\) 1 part = 50mL. Milk=350, Cream=100.
  2. New: Ratio 5:3. Total still 450mL. 8 parts = 450mL.
  3. 1 part = \(450/8 = 56.25\).
  4. New Milk = \(5 \times 56.25 = 281.25\).
  5. Difference = \(350 - 281.25 = 68.75\) mL.

Percents and Financial Math

This section is referenced in the table of contents but detailed content is not provided in the original document.

3 Algebra & Logic

Sequences, Recursion, and Functions

This section is referenced in the table of contents but detailed content is not provided in the original document.

4 Geometry

Angles and Polygons

Key Ideas & When to Use

  • Trigger: "Star polygon," "Interior angles," "Zig-zag lines."
  • Formulas:
    • Sum of Interior Angles: \((n-2) \times 180^\circ\).
    • Sum of Exterior Angles: Always \(360^\circ\).
    • Regular Polygon Angle: \(\frac{(n-2)180}{n}\).

Triangles: Centers and Areas

Key Ideas & When to Use

  • Trigger: "Inscribed circle," "Medians," "Altitudes."
  • Triangle Centers:
    1. Centroid (G): Intersection of medians. Divides median 2:1.
    2. Incenter (I): Intersection of angle bisectors. Center of Incircle. \(Area = r \times s\).
    3. Circumcenter (O): Intersection of perpendicular bisectors. Center of Circumcircle.
    4. Orthocenter (H): Intersection of altitudes.

Mini Toolkit (Formulas)

  • Heron's Formula: \(A = \sqrt{s(s-a)(s-b)(s-c)}\).
  • Inradius: \(A = rs\) (\(s\) is semi-perimeter).
  • Circumradius: \(R = \frac{abc}{4A}\) (or \(R = c/2\) for right triangles).
  • Coordinates: Shoelace Theorem for area given \((x,y)\) coordinates.
Representative Example

Problem: 2004 AMC 10B #22 (Adapted for AMC 8)
Triangle sides 5, 12, 13. Find distance between center of incircle and circumcircle.

Solution:
  1. Right triangle (5-12-13). Area = 30.
  2. Inradius: \(r = \frac{A}{s} = \frac{30}{(5+12+13)/2} = \frac{30}{15} = 2\).
  3. Circumradius (Right Triangle): \(R = \frac{\text{hypotenuse}}{2} = 6.5\).
  4. Distance (Euler's Theorem): \(d^2 = R(R-2r)\).
  5. \(d^2 = 6.5(6.5 - 4) = 6.5(2.5) = 16.25\).
  6. \(d = \sqrt{16.25} = \frac{\sqrt{65}}{2}\).

Circles, Tangents, and Power of a Point

Key Ideas & When to Use

  • Trigger: Lines intersecting circles, tangent lines from a point.
  • Theorem: Tangent \(\perp\) Radius.
  • Two-Tangent Theorem: Tangents from an external point are equal length (\(PA=PB\)).
  • Power of a Point:
    • Secant-Secant: \(PA \cdot PB = PC \cdot PD\).
    • Tangent-Secant: \(PT^2 = PA \cdot PB\).
Representative Example

Problem: 2018 AMC 8 #15
Two small circles diameter lies on radius of large circle. Combined area 1. Area of shaded?

Solution:
  1. Let small radius = \(r\), large radius \(R = 2r\).
  2. Area small = \(\pi r^2\). Combined = \(2\pi r^2 = 1\).
  3. Area large = \(\pi (2r)^2 = 4\pi r^2 = 2(2\pi r^2) = 2(1) = 2\).
  4. Shaded = Large - Small = \(2 - 1 = 1\).

5 Counting & Probability

Systematic Counting & PIE

Key Principles

  • Principle of Inclusion-Exclusion (PIE): \(|A \cup B| = |A| + |B| - |A \cap B|\).
  • Complementary Counting: Count what you don't want and subtract from Total.

Advanced Counting: Stars & Bars and Catalan

Key Ideas & When to Use

  • Stars and Bars Trigger: "Distribute \(n\) identical candies to \(k\) distinct children."
    • Formula (Non-negative): \(\binom{n+k-1}{k-1}\).
    • Formula (Positive/At least 1): \(\binom{n-1}{k-1}\).
  • Catalan Numbers Trigger: "Balanced parentheses," "Mountain ranges," "Triangulating a polygon," "2xN grid paths not crossing diagonal."
    • Sequence: 1, 1, 2, 5, 14, 42...
    • Formula: \(C_n = \frac{1}{n+1}\binom{2n}{n}\).
Representative Example

Problem: Number of ways to triangulate a hexagon.

Solution:
  1. A hexagon has \(n=6\) vertices.
  2. Triangulation of polygon with \(n+2\) sides is \(C_n\).
  3. Here sides = 6, so we need \(C_4\).
  4. \(C_4 = \frac{1}{5}\binom{8}{4} = \frac{1}{5} \times 70 = 14\).

Probability & Recursion

Key Ideas & When to Use

  • Trigger: "Probability after \(n\) steps," "Hopping cricket."
  • Core Method: Create a state table or tree diagram. Look for a pattern in the probability or count.
Representative Example

Problem: 2022 AMC 8 #25
Cricket hops on 4 leaves. Starts at leaf S. 4 hops. Probability of returning to S?

Solution (Recursive):
  1. Let \(S_n\) be ways to be at Start, \(O_n\) be ways to be at Other (3 leaves).
  2. Start: \(S_0 = 1, O_0 = 0\).
  3. Hop 1: Must go to Other. \(S_1 = 0, O_1 = 3\).
  4. Hop 2: From Other, can go back to S (1 way) or other O (2 ways).
    • \(S_2 = O_1 \times 1 = 3\).
    • \(O_2 = O_1 \times 2 = 6\).
  5. Hop 3:
    • \(S_3 = O_2 \times 1 = 6\).
    • \(O_3 = S_2 \times 3 + O_2 \times 2 = 3(3) + 6(2) = 9 + 12 = 21\).
  6. Hop 4:
    • \(S_4 = O_3 \times 1 = 21\).
  7. Total paths = \(3^4 = 81\).
  8. Probability = \(21/81 = 7/27\).