Competition Math Secrets: Tricks the Top Students Know

Math competitions like AMC, MATHCOUNTS, and math olympiads test more than memorized formulas. They test problem-solving instinct — the ability to look at an unfamiliar problem and find the clever shortcut hiding inside it.

The good news? These “clever shortcuts” are actually learnable techniques. Here are the tricks that top competitors reach for again and again.

This is Part 7 (the finale!) of our “Math Tricks Your Textbook Never Taught You” series.

1. The Chicken McNugget Theorem

What it solves: “What's the largest number you can't make?”

If you can only buy nuggets in packs of a and b (where a and b share no common factor), the largest number of nuggets you cannot buy is:

ab − a − b

Example: Packs of 3 and 5.

Largest impossible count = 3 × 5 − 3 − 5 = 7
Check: you can't make 7 from 3s and 5s. But 8 = 3 + 5, 9 = 3 + 3 + 3, 10 = 5 + 5 — every number above 7 works.

Example: Packs of 7 and 11.

Largest impossible count = 7 × 11 − 7 − 11 = 77 − 18 = 59

This appears on AMC/MATHCOUNTS nearly every year in some form. Any time a problem says “two coprime denominations,” reach for this formula.

2. Stars and Bars

What it solves: Counting ways to distribute identical items into groups.

How many ways can you put n identical balls into k distinct boxes? Imagine the balls as stars and the dividers between boxes as bars:

★★★ | ★ | ★★ = (3, 1, 2)

You need n stars and k − 1 bars, and you're choosing where to place the bars:

C(n + k − 1, k − 1)

Example: Distribute 10 candies among 3 children.

n = 10 candies, k = 3 children
Answer = C(10 + 3 − 1, 3 − 1) = C(12, 2) = 66 ways

Example: How many solutions does x + y + z = 8 have (non-negative integers)?

Same formula: C(8 + 3 − 1, 2) = C(10, 2) = 45

3. The Pigeonhole Principle

What it solves: Proving something must happen.

If you stuff n + 1 pigeons into n holes, at least one hole has at least 2 pigeons. Sounds obvious, but it's incredibly powerful.

Example: In a room of 13 people, prove at least 2 share a birth month.

13 people (pigeons) → 12 months (holes) → at least one month has ≥ 2 people. Done.

Example: Pick any 5 integers. Prove that 2 of them have the same remainder when divided by 4.

Only 4 possible remainders: 0, 1, 2, 3 (the “holes”).
5 integers (pigeons) → 4 holes → at least 2 share a remainder.

Generalized version: If kn + 1 pigeons go into n holes, at least one hole has k + 1 pigeons.

4. Modular Arithmetic (Clock Math)

What it solves: Remainder problems, last-digit problems, divisibility.

Think of numbers on a clock. On a 12-hour clock, 15 o'clock = 3 o'clock. In math notation: 15 ≡ 3 (mod 12).

Last digit trick: What's the last digit of 7¹⁰⁰?

7¹ = 7, 7² = 49, 7³ = 343, 7⁴ = 2401
Last digits cycle: 7, 9, 3, 1, 7, 9, 3, 1… (period 4)
100 ÷ 4 = 25 remainder 0, so 7¹⁰⁰ has the same last digit as 7⁴ = 1

Remainder trick: What's the remainder when 2²⁰ is divided by 7?

2¹ ≡ 2, 2² ≡ 4, 2³ ≡ 1 (mod 7)
Since 2³ ≡ 1 (mod 7), then 2²⁰ = (2³)⁶ × 2² ≡ 1⁶ × 4 = 4

5. Telescoping

What it solves: Sums that look impossible but collapse like a telescope.

The key idea: rewrite each term as a difference, then watch most terms cancel.

Example: Find 1/(1×2) + 1/(2×3) + 1/(3×4) + … + 1/(99×100)

Rewrite: 1/(n(n+1)) = 1/n − 1/(n+1) (partial fractions)
Sum becomes: (1/1 − 1/2) + (1/2 − 1/3) + (1/3 − 1/4) + … + (1/99 − 1/100)
Everything cancels except the first and last: 1 − 1/100 = 99/100

Another example: Find 1×2 + 2×3 + 3×4 + … + n(n+1)

Note: n(n+1) = [n(n+1)(n+2) − (n−1)n(n+1)] / 3
Telescopes to: n(n+1)(n+2)/3
For n = 10: 10 × 11 × 12 / 3 = 440

6. Simon's Favorite Factoring Trick (SFFT)

What it solves: Equations with two variables multiplied together.

When you see xy + ax + by = c, add ab to both sides so you can factor:

xy + ax + by + ab = c + ab
(x + b)(y + a) = c + ab

Now you just need factor pairs of the right side!

Example: Find positive integers where xy + 3x + 2y = 29.

Add 3 × 2 = 6 to both sides: xy + 3x + 2y + 6 = 35
Factor: (x + 2)(y + 3) = 35
Factor pairs of 35: (1, 35), (5, 7), (7, 5), (35, 1)
So (x + 2, y + 3) can be (5, 7) or (7, 5) → (x, y) = (3, 4) or (5, 2)

Example: Solve xy − 2x − 3y = 7 in positive integers.

Add (−2)(−3) = 6 to both sides: xy − 2x − 3y + 6 = 13
Factor: (x − 3)(y − 2) = 13
13 is prime: (1, 13) or (13, 1)
(x − 3, y − 2) = (1, 13) → (x, y) = (4, 15) or (13, 1) → (16, 3)

7. Vieta's Formulas

What it solves: Finding sums and products of roots without solving the equation.

For a quadratic ax² + bx + c = 0 with roots r and s:

r + s = −b/a
r × s = c/a

Example: If r and s are roots of x² − 7x + 12 = 0, find r² + s².

r + s = 7, rs = 12 (by Vieta's)
r² + s² = (r + s)² − 2rs = 49 − 24 = 25
(No need to find r = 3, s = 4 separately!)

For cubics x³ + bx² + cx + d = 0 with roots r, s, t:

r + s + t = −b
rs + rt + st = c
rst = −d

8. Inclusion-Exclusion Principle

What it solves: Counting with overlapping sets.

For two sets: |A ∪ B| = |A| + |B| − |A ∩ B|

For three sets: |A ∪ B ∪ C| = |A| + |B| + |C| − |A ∩ B| − |A ∩ C| − |B ∩ C| + |A ∩ B ∩ C|

Example: Of 100 students, 60 like pizza, 45 like burgers, and 20 like both. How many like neither?

Like at least one = 60 + 45 − 20 = 85
Like neither = 100 − 85 = 15

Classic competition problem: How many integers from 1 to 1000 are divisible by neither 2, 3, nor 5?

|div by 2| = 500, |div by 3| = 333, |div by 5| = 200
|div by 6| = 166, |div by 10| = 100, |div by 15| = 66
|div by 30| = 33
By inclusion-exclusion: 500 + 333 + 200 − 166 − 100 − 66 + 33 = 734
Neither: 1000 − 734 = 266

9. The Hockey Stick Identity

What it solves: Summing entries along a diagonal of Pascal's triangle.

The identity says: the sum of entries going diagonally down Pascal's triangle equals the entry one step further “down and to the side” — like the blade of a hockey stick.

C(r,r) + C(r+1,r) + C(r+2,r) + … + C(n,r) = C(n+1, r+1)

Example: What's C(2,2) + C(3,2) + C(4,2) + C(5,2) + C(6,2)?

= 1 + 3 + 6 + 10 + 15 = 35
By hockey stick: C(7, 3) = 35. Confirmed!

10. Pick's Theorem

What it solves: Area of a polygon drawn on a grid (lattice points).

Area = I + B/2 − 1

Where I = interior lattice points and B = boundary lattice points.

Example: A triangle with vertices at (0,0), (4,0), and (0,3).

Boundary points: 5 on bottom edge, 4 on left edge, 2 on hypotenuse (plus vertices) = 5 + 4 + 1 = 10. Wait — let's count carefully:
Bottom: (0,0), (1,0), (2,0), (3,0), (4,0) = 5 points
Left: (0,1), (0,2), (0,3) = 3 points (not counting origin again)
Hypotenuse from (4,0) to (0,3): GCD(4,3) + 1 = 2 interior + endpoints, but endpoints already counted. GCD(4,3) − 1 = 0 new points.
B = 5 + 3 + 0 = 8 boundary points
Interior points: (1,1), (2,1), (1,2) = 3 points
Area = 3 + 8/2 − 1 = 3 + 4 − 1 = 6
Check: ½ × 4 × 3 = 6. Confirmed!

11. Complementary Counting

What it solves: “At least one” problems.

Instead of counting what you want directly, count what you don't want and subtract:

What we want = Total − What we don't want

Example: Roll two dice. What's the probability of getting at least one 6?

Total outcomes = 36
No sixes: 5 × 5 = 25
At least one six: 36 − 25 = 11
Probability = 11/36

Example: How many 3-digit numbers have at least one even digit?

Total 3-digit numbers: 900 (from 100 to 999)
All digits odd: 5 choices for first digit (1,3,5,7,9) × 5 × 5 = 125
At least one even digit: 900 − 125 = 775

12. Difference of Squares (Competition Style)

What it solves: Quick factoring of numbers on timed tests.

You know a² − b² = (a+b)(a−b). Competition students use it in reverse to do instant arithmetic:

Example: Is 1,000,003 prime?

1,000,003 = 1,000,000 + 3 — not an obvious form.
But: 1003² = 1,006,009 and 1002² = 1,004,004. Not helpful directly.
Try: 1,000,003 = 1,000,000 + 3 = 10⁶ + 3. Not a difference of squares.
Actually: 1,000,003 = 1003 × 997 − 3 + ... Let's just use the trick for easier cases.

Better example: Calculate 47 × 53 instantly.

= (50 − 3)(50 + 3) = 50² − 3² = 2500 − 9 = 2491

Factor 9,991:

9,991 = 10,000 − 9 = 100² − 3² = (100 + 3)(100 − 3) = 103 × 97

13. The Invariant Strategy

What it solves: Proving something is impossible or must be true.

Find a quantity that never changes (an invariant), no matter what moves are made. If the starting and ending states have different invariant values, the transformation is impossible.

Classic example: A checkerboard has two opposite corners removed. Can you tile the remaining 62 squares with 31 dominoes?

Each domino covers exactly one black and one white square.
The two removed corners are the same color (both white, on a standard board).
Remaining: 30 white + 32 black (or vice versa).
31 dominoes need 31 white and 31 black, but we have 30 and 32.
Impossible! The parity invariant proves it.

14. The Extreme Principle

What it solves: Existence proofs and construction problems.

Look at the largest, smallest, leftmost, or most extreme element. It often has special properties that crack the problem open.

Example: Given n points in the plane, not all on one line, prove there exists a line passing through exactly 2 of them.

Consider all pairs (point, line through 2+ points). Pick the pair where the point-to-line distance is smallest.
That line must pass through exactly 2 points (if it passed through 3, you could find a closer point-line pair, contradiction).

How to Build Competition Problem-Solving Skills

Recognize the pattern. Most problems fall into one of these categories. When you read a problem, ask: “Is this pigeonhole? Complementary counting? Modular arithmetic?”
Practice with purpose. Don't just solve problems — after solving, ask “What was the key insight?” and categorize the technique.
Build a toolkit. Keep a notebook of techniques with one example each. Review before competitions.
Work backwards. If you're stuck, ask “What would the answer look like?” and work from there.
Try small cases. If a problem asks about 100, try it with 3, 4, 5 first. Patterns emerge.

This is Part 7 (the finale!) of our “Math Tricks Your Textbook Never Taught You” series. If you missed any, start from the beginning with Visual Multiplication Tricks From Around the World.