Reaching for a perfect score — the climb to 20+

The 20+ problems lean on exponent moves done in your head, not on a single rule. Carry these:

• Read every exponent as a copy-count so the laws are obvious: same base → multiply adds powers \(a^m \cdot a^n = a^{m+n}\), divide subtracts \(a^m \div a^n = a^{m-n}\), power-of-a-power multiplies \((a^m)^n = a^{mn}\).
• The staircase fixes the edge cases without memorizing: \(a^0 = 1\) and \(a^{-n} = 1/a^n\) are just steps below \(a^1\), each dividing by the base.
• Square big numbers fast with \((a + b)^2 = a^2 + 2ab + b^2\): e.g. \(21^2 = 20^2 + 2\cdot 20 + 1 = 441\). The middle \(2ab\) is the piece most kids drop.
• Scientific notation turns ugly products into magnitude bookkeeping: \((a \times 10^m)(b \times 10^n) = ab \times 10^{m+n}\) — multiply the fronts, add the exponents, count zeros.

A 19th-century shortcut for checking arithmetic without redoing it. It catches most slips in seconds.

The fact: a number and its digit sum leave the same leftover when divided by 9. (Why? 10, 100, 1000, … all leave leftover 1 when divided by 9, so each digit contributes only itself.)

How to use it. Compute your answer, then check:

1. Take the digit sum of each input; keep adding digits until one digit is left.
2. Combine those single digits the same way the problem combines the inputs (+ if you added, × if you multiplied).
3. Take the digit sum of your computed answer.
4. The two single digits should match. If they do not, you slipped.

Try it. Is 472 + 351 = 823? Digit sums: 4+7+2 = 13 → 4; 3+5+1 = 9 → 0 (since 9 acts like 0 here). Combine: 4 + 0 = 4. Digit sum of 823: 8+2+3 = 13 → 4. ✓ They match, so the answer is consistent.

It does not catch every error — swapping two digits leaves the digit sum unchanged — but it catches most, fast.

• Perfect-square filter (mod 3, mod 4). A perfect square can ONLY have these remainders:
   · mod 3: 0 or 1 (never 2). So if N ≡ 2 (mod 3), N is NOT a perfect square.
   · mod 4: 0 or 1 (never 2 or 3). So if N ≡ 2 or 3 (mod 4), N is NOT a perfect square.
  Great for “is X a perfect square?” questions — rules out most numbers in one glance.
• Product of divisors of n: nd/2 where d is the number of divisors. (Pair each divisor with its partner; the product of every pair is n — the same pairing idea behind the divisor-sum trick in chapter 4.)
• Chicken McNugget Theorem. If a and b are coprime positive integers, the largest amount you CANNOT make as a non-negative combination of a’s and b’s is ab − a − b. (For 4 and 7: the largest unmakeable amount is 4·7 − 4 − 7 = 17.)
• Legendre’s formula (exponent of prime p in n!): vp(n!) = ⌊n/p⌋ + ⌊n/p²⌋ + ⌊n/p³⌋ + … — useful for “how many trailing zeros does 100! have?” (answer: v₅(100!) = 20 + 4 = 24).

• Two-apart split: 1 / (n · (n+2)) = ½[1/n − 1/(n+2)]. So 1/(1·3) + 1/(3·5) + 1/(5·7) + … telescopes too, now with a 1/2 out front.
• Product telescoping with (1 + 1/k): (1 + 1/1)(1 + 1/2)(1 + 1/3) … (1 + 1/n) = (2/1)(3/2)(4/3) … ((n+1)/n) = n + 1. Each numerator kills the next denominator. The whole product collapses to n + 1.
• The harmonic series. 1 + 1/2 + 1/3 + … does NOT telescope — it grows without bound (slowly). Don’t try.
• Any repeating decimal → fraction. Let x equal the decimal, multiply by a power of 10 to slide one whole period, subtract: 0.3 → 10x−x = 3 → x = 1/3. For a partly-repeating decimal, slide so the loops line up before subtracting: 0.28 → 10x−x = 2.6 → x = 26/90 = 13/45.
• Repeat-block length. For 1/d (in lowest terms, no factor of 2 or 5 in d), the repeating block has length at most d−1 — the remainders cycle: 1/7 = 0.142857, a block of 6.
• Mixed-number arithmetic fast. For a single hard subtraction, flip both to improper over a common bottom (no borrowing). For a long sum, estimate: each mixed number is its whole part plus a bit under 1, so the total is a little above the sum of the whole parts — often enough to pick the smallest whole number above it.

• Pick convenient numbers. When only the shape of a trip is given (fractions, percents, no real distance/time), choose the total yourself to kill the denominators — thirds → use 3, quarters → use 4. The answer can't depend on the real size.
• Round-track repeated meetings. Opposite ways: they meet every track ÷ (v₁ + v₂). Same way: the faster laps the slower every track ÷ (vfast − vslow).
• Three-worker together-time (alone a, b, c): T = 1 / (1/a + 1/b + 1/c). Same pattern, three terms.
• Filling AND draining at once. Fill rate 1/a, drain rate 1/d give net rate 1/a − 1/d — the drain subtracts.
• Mixture problems. Mixing x liters of A% with y liters of B% gives concentration (xA + yB) / (x + y) — a weighted average.
• Train passes a person vs. a station. Past a point: train travels its own length. Past a station: train length + station length. Time = length ÷ train speed.
• Continuous compound interest (rare here): ert grows faster than discrete (1+r)^t for the same r, t.

• Symmetric-sum trick. When several equations share a symmetric structure (e.g., 2x + y + z = a, x + 2y + z = b, x + y + 2z = c), ADD them all. The left side simplifies to a clean multiple of (x + y + z). Solve for the symmetric sum first; then individual variables.
• Sum & product of two numbers. If x + y = S and xy = P, then x and y are the roots of t² − St + P = 0. So x² + y² = S² − 2P. Useful when the problem gives you S and P without giving x and y individually.
• Simon’s Favorite Factoring Trick (SFFT). An equation like xy + 3x + 2y = 17 can be tamed by adding a constant to both sides to factor: xy + 3x + 2y + 6 = 23 ⇒ (x+2)(y+3) = 23. Now hunt integer factor pairs of 23.
• Sum/difference of cubes: a³ + b³ = (a + b)(a² − ab + b²); a³ − b³ = (a − b)(a² + ab + b²).
• Reduce and Expand (for scary numbers). Faced with 'the 100th figure' or 'a 26-team league', don't fight the big number — shrink it to the first 4–5 cases, tabulate, and read the first differences between terms. If the terms have no fixed step, the jumps usually do (triangular numbers 1, 3, 6, 10 jump by 2, 3, 4, 5). Find the rule small, then grow it back.

• British Flag Theorem. For any point P inside a rectangle ABCD: PA² + PC² = PB² + PD². Squared distances to opposite corners stay equal — wherever P sits.
• Euler’s polyhedron formula. For any flat-faced solid: V − E + F = 2. Cube: 8 − 12 + 6 = 2 ✓. Tetrahedron: 4 − 6 + 4 = 2 ✓.
• Inradius of a right triangle (legs a, b, hypotenuse c): r = (a + b − c) / 2. The circle in a 3-4-5 triangle has radius (3 + 4 − 5)/2 = 1.
• Heron’s formula (area from sides only): with s = (a+b+c)/2, area = √(s(s−a)(s−b)(s−c)). Avoid unless you must — a height or a decomposition is usually cleaner.

• PIE for 3 sets: |A ∪ B ∪ C| = |A| + |B| + |C| − |A∩B| − |A∩C| − |B∩C| + |A∩B∩C|.
• Pascal’s identity: C(n, k) + C(n, k+1) = C(n+1, k+1). (Useful for building Pascal’s triangle row-by-row.)
• Binomial sum: C(n,0) + C(n,1) + … + C(n,n) = 2ⁿ. (Same fact as “number of subsets” — two views.)
• Linearity of expectation: E[X + Y] = E[X] + E[Y], even when X and Y are NOT independent. Lets you compute the expected total without disentangling cases.
• Recursion-as-counting: when a problem builds on smaller cases, count f(1), f(2), f(3) and look for a relation like f(n) = f(n−1) + f(n−2) (Fibonacci). Then chase the recurrence.
• Count two ways / invariants: find one total that two different counts must both equal and let them pin each other down — e.g. sum of handshake counts = 2×(handshakes), so the number of odd-degree people is even (degree-sum parity); or “count the losers” in a knockout (n−1 games).

• Extremal answers need TWO halves. For ‘least possible largest’ or ‘greatest possible least’, pair a lower bound (a counting reason it can’t do better) with a construction (an actual example that hits the bound). One half alone never settles it.
• Liar puzzles at scale — count, don’t case. When dozens of people make claims around a table, stop branching and count: honest speakers must all agree on a shared fact, so different answers cap how many can be honest, and the liars’ fixed values often pin the true total.
• Reduce-and-expand to a formula. Push the small cases far enough to name the rule, then prove it: pairwise connections give \(\dfrac{n(n-1)}{2}\); ‘last move’ counting gives recurrences like \(f(n)=f(n-1)+f(n-2)\). A guessed pattern is a #25 trap; a proven one is a tool.
• Cycles beat brute force. Units digits, weekdays, and repeating sequences all reduce a giant position to (position mod L) — and remainder 0 lands on the LAST slot. Verify the cycle length on a tiny case before you trust it.
• Pigeonhole for ‘must’ questions. ‘How many must you take to guarantee …’ is solved by filling the worst case to the brim, then adding one more.