Lesson 2: Arithmetic Mastery

The Gaokao forbids calculators. Every mark you earn is earned by hand, which makes fluent, accurate arithmetic a competitive weapon rather than a primary-school topic. Chinese mathematics has always prized this: Sunzi's 算经 taught systematic computation in the 3rd century, and the abacus (算盘) trained generations of clerks to out-calculate machines. This lesson sharpens the four operations on signed numbers, percentages, estimation, and — most importantly — the verification habits that separate a correct answer from a near miss. In examination statistics, careless arithmetic errors cost more marks than hard problems do.

Order of operations and signed numbers

The hierarchy: brackets first, then powers, then multiplication/division (left to right), then addition/subtraction (left to right). Two traps dominate. First, −3² = −(3²) = −9, while (−3)² = +9: the power binds to what is inside the bracket only. Second, dividing by a negative flips signs at every step: (−24) ÷ (−4) = 6, but (−24) ÷ 4 = −6. Write one operation per line; mental compression is where signs die.

Worked examples

Worked example 1 — signed arithmetic. Evaluate 18 ÷ (−3)² − (−2)³ × (1/2). Step 1, powers: (−3)² = 9 and (−2)³ = −8. Step 2, division and multiplication: 18 ÷ 9 = 2; (−8) × (1/2) = −4. Step 3, subtraction: 2 − (−4) = 6. Estimate first: the first term is small and positive, the second term is negative and subtracting it adds — an answer near 6 is plausible; an answer of −2 would signal a sign error.

Worked example 2 — speed through structure. (a) 999 × 37 + 37. Factor the hidden 37: 37 × (999 + 1) = 37,000. (b) 25 × 36: regroup as 25 × 4 × 9 = 100 × 9 = 900. (c) 0.125 × 32 × 2.5: split 32 = 8 × 4 and pair 0.125 × 8 = 1 with 4 × 2.5 = 10, giving 10. The distributive and associative laws are not abstractions — they are how fast calculators think. Hunt for 25s and 125s and their partners (×4, ×8) in every product.

Worked example 3 — successive percentage changes. A jacket costs ¥800. The shop discounts 20%, then a coupon takes a further 10% off the discounted price. Price after the first discount: 800 × 0.8 = ¥640. After the coupon: 640 × 0.9 = ¥576. The total discount is (800 − 576)/800 = 28%, not 30%. Percentages multiply; they do not add, because the second percentage acts on a smaller base. The same warning applies to "up 10% then down 10%": 1.1 × 0.9 = 0.99 — you end 1% down, not level.

Worked example 4 — estimation with units. A car uses 7.8 L of fuel per 100 km. Estimate the fuel for a 460 km trip, and its cost at ¥7.6 per litre. Round 7.8 ≈ 8 and split 460 into 4.6 hundreds: fuel ≈ 8 × 4.6 ≈ 36.8, about 36 L. Cost ≈ 36 × 7.6 ≈ ¥274. (Exact: 35.88 L and ¥272.69 — the estimate is within 1%.) Estimation also polices units: litres per 100 km times hundreds of kilometres gives litres; if your manipulation produces ¥ per km, a unit was dropped.

Challenge problem — Gaokao standard

Problem. Compute 2² − 1² + 4² − 3² + 6² − 5² + … + 100² − 99².

Solution. Pair consecutive terms and use the difference of squares: (2k)² − (2k−1)² = (2k + (2k−1)) × (2k − (2k−1)) = (4k − 1) × 1 = 4k − 1, for k = 1, 2, …, 50. The sum becomes Σ(4k − 1) = 4 × (1 + 2 + … + 50) − 50 × 1. The Gauss sum 1 + … + 50 = 50 × 51/2 = 1275, so the total is 4 × 1275 − 50 = 5100 − 50 = 5050. Brute force would take a hundred multiplications; structure takes four lines. (Amusing check: the answer equals 1 + 2 + … + 100.)

Common mistakes

• −3² computed as 9. Without brackets, the square applies to 3 and the minus stands outside: −3² = −9.
• Adding successive percentages. 20% off then 10% off is 28% off (Worked example 3). Percent changes compose multiplicatively.
• Percentage points versus percent. A rate rising from 4% to 5% rose by 1 percentage point but by 25% relatively. News reports exploit this confusion deliberately.
• Rounding mid-calculation. Round only at the end, or keep two guard digits. Rounding √5 to 2 too early would destroy an integer-part argument.
• Skipping the estimate. An answer of ¥27,000 for the fuel in Worked example 4 is a misplaced decimal — visible instantly if you estimated first.

Exam Experience

Multiple choice. The value of (−1)²⁰²⁵ + |−3| − ∛(−8) is: A. −4 B. 0 C. 4 D. 6 Answer: C. (−1) to an odd power is −1; |−3| = 3; ∛(−8) = −2, and subtracting it adds 2. Total: −1 + 3 + 2 = 4.

Fill in the blank. A price is raised by 10% and later reduced by 10%. The final price is ____ % of the original. Answer: 99. The multiplier is 1.1 × 0.9 = 0.99.

Solution question (10 marks). A city's taxi fare is ¥10 for the first 3 km; beyond 3 km the charge is ¥2.4 per kilometre on the exact distance travelled. (1) Find the fare for a 12.5 km ride. (2) A passenger has ¥50. What is the greatest whole number of kilometres they can travel? Marking scheme. (1) Distance beyond 3 km: 12.5 − 3 = 9.5 km — 2 marks. Charge: 9.5 × 2.4 = ¥22.8 — 2 marks. Total fare 10 + 22.8 = ¥32.8 — 1 mark. [5 marks] (2) Money available for the per-kilometre part: 50 − 10 = ¥40 — 1 mark. Distance it buys: 40 ÷ 2.4 = 16.66… km — 2 marks. Total distance 3 + 16.66… = 19.66… km, so the greatest whole number of kilometres is 19 km — 2 marks for rounding DOWN (rounding to 20 would exceed the budget). [5 marks] Total: 10 marks. The last mark is for interpreting the number, not computing it — applied questions end with a judgement.

Going deeper — casting out nines

Before calculators, clerks checked long multiplication by casting out nines: a number and the sum of its digits leave the same remainder on division by 9, because 10 ≡ 1 (mod 9) makes every power of 10 collapse to 1. Check 487 × 36 = 17,532. Digit sums: 487 → 4 + 8 + 7 = 19 → 1 (mod 9); 36 → 9 → 0. Product of remainders: 1 × 0 = 0. Candidate answer: 1 + 7 + 5 + 3 + 2 = 18 → 0 ✓ — consistent (and the multiplication is indeed correct). The check is necessary, not sufficient: it cannot catch an error that is a multiple of 9, most famously transposed digits (17,352 also passes). Chinese abacus masters used it as a first-line audit; treat it the same way — a fast filter that catches most slips, layered under Lesson 2's estimate-first habit and Lesson 20's independent-route verification.

The same congruence explains the divisibility rules for 3 and 9: a number is divisible by 3 (or 9) exactly when its digit sum is. Test 123,456: digit sum 21 — divisible by 3 (21 = 3 × 7) but not by 9. Indeed 123,456 = 3 × 41,152, and 41,152's digit sum is 13, not divisible by 3, so the exact power of 3 dividing 123,456 is 3¹. Divisibility by 11 has its own gem: alternate adding and subtracting digits (since 10 ≡ −1 mod 11). For 123,456: 6 − 5 + 4 − 3 + 2 − 1 = 3 — not divisible by 11. These are not party tricks; they are modular arithmetic (Lesson 14's indices, Lesson 17's invariants) in everyday dress, and the Gaokao's number-theory-flavoured fill-ins reward students who see them.

Second look — a wholesaler's ledger, audited three ways

A wholesaler buys 250 kg of tea at ¥48 per kg, pays ¥350 for transport, and sells at ¥56 per kg — but 5% of the stock spoils before it can be sold. Find the profit, and audit each computation.

Cost of stock: 250 × 48. Compute structurally (Lesson 2): 25 × 48 = 1200, so 250 × 48 = ¥12,000. Casting out nines: 250 → 7, 48 → 3, product of remainders 21 → 3; and 12,000 → 1 + 2 = 3 ✓ consistent. Total cost: 12,000 + 350 = ¥12,350. Sellable stock: 5% of 250 kg is 12.5 kg spoiled, leaving 237.5 kg. Revenue: 237.5 × 56. Compute: 237.5 × 56 = 237.5 × 50 + 237.5 × 6 = 11,875 + 1,425 = ¥13,300. Estimate audit: roughly 240 × 55 ≈ ¥13,200 — the right neighbourhood ✓. Profit: 13,300 − 12,350 = ¥950.

Now the instructive failure: a clerk computes 250 × 48 = ¥11,200 (a slip: 25 × 48 mis-multiplied). Which audits catch it? The estimate (250 × 48 ≈ 250 × 50 = 12,500, and must exceed 12,000 since 48... more precisely 48 > 40 forces the product above 10,000, but the sharper structural route 25 × 48 = 1200 exposes it instantly). Casting out nines: 11,200 → digit sum 4, but the true remainder must be 3 — caught. This is why professionals layer checks: structure first, nines second, estimate always. An error that survives one independent check rarely survives two.

One last drill — exact division with an audit. Compute 2025 ÷ 45. Estimate first: 45 × 40 = 1800 and 45 × 50 = 2250, so the quotient lies between 40 and 50. Try 45: 45 × 45 = 2025 exactly — the quotient is 45. Audit by casting out nines: 2025 → 9 → 0 (mod 9); 45 → 0; 0 × 0 = 0 ✓ consistent. A perfect-square coincidence (45² = 2025) hides inside a routine division; the habit of estimating before dividing is what positions you to notice it. File this under Lesson 2's whole philosophy: arithmetic performed with attention is arithmetic performed once.