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📅 Today's Problem
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📋The 4-Step Framework — expand for guiding prompts▼
STEP 1
Understand the Problem
🔍
What is the unknown?
What is given?
What is the condition?
Can you draw a figure?
Restate it in your own words
STEP 2
Devise a Plan
🗺️
Seen a similar problem?
Solve a simpler version?
Split the conditions?
Add an auxiliary element?
Work backwards from the answer?
STEP 3
Carry Out the Plan
✏️
Can you justify each step?
Where exactly does it break?
Is the premise valid?
Algebra error or wrong plan?
Try a different variable?
STEP 4
Look Back
🔄
Is the result reasonable?
Can you verify it another way?
Can you generalize it?
Is there a simpler solution?
Add it to your error log
📖Classic Problems — see the 4 steps in action▼
Western ClassicsChinese Classics 🏮
Problem (US classic): You have 30 coins made up of quarters ($0.25) and dimes ($0.10). The total value is $5.10. How many quarters and how many dimes do you have?
1
Understand the Problem
Unknown: # quarters (q), # dimes (d). Given: q + d = 30 coins; 0.25q + 0.10d = 5.10 dollars. → Two unknowns, two equations — this is a system.
2
Devise a Plan
Assume all 30 are dimes: 30 × $0.10 = $3.00. Actual value is $5.10 — gap of $2.10. Each swap (dime → quarter) adds $0.15. → # quarters = $2.10 ÷ $0.15.
Check: 14 × $0.25 + 16 × $0.10 = $3.50 + $1.60 = $5.10 ✓; 14 + 16 = 30 ✓ → Same trick as any "two types, two constraints" mixture problem. The "assume all one type" strategy is a universal key.
Problem (Work Rate — classic US/UK curriculum): Alice can paint a fence in 4 hours. Bob can paint the same fence in 6 hours. How long does it take them working together?
1
Understand the Problem
Unknown: time t (hours) working together. Given: Alice's rate = 1/4 fence per hour; Bob's rate = 1/6 fence per hour. Condition: rates add when working simultaneously.
2
Devise a Plan
Think in rates, not time. In 1 hour together they complete: 1/4 + 1/6 of the fence. Time = Total work ÷ Combined rate = 1 ÷ (1/4 + 1/6).
Is 2.4 hrs reasonable? Must be less than the faster worker (4 hrs). 2.4 < 4 ✓ Verify: Alice paints 2.4/4 = 0.6; Bob paints 2.4/6 = 0.4. Total = 1.0 ✓ → Works for pipes, printers, assembly lines — any "combined rate" scenario.
Problem (Optimization — US Algebra / Pre-Calc standard): A farmer has 120 feet of fencing. He wants to fence a rectangular garden against a barn wall (the barn forms one long side — no fence needed there). What dimensions give the maximum area?
1
Understand the Problem
Unknown: width w and length l (parallel to barn). Given: only 3 sides need fencing → 2w + l = 120. Goal: maximize area A = l × w. → Draw the rectangle with barn as one side.
2
Devise a Plan
From constraint: l = 120 − 2w. Substitute into area: A = (120 − 2w) × w = 120w − 2w² This is a downward parabola → max at the vertex.
3
Carry Out the Plan
A = −2(w² − 60w) = −2(w − 30)² + 1800 Vertex at w = 30 ft l = 120 − 2(30) = 60 ft Max area = 30 × 60 = 1800 sq ft
4
Look Back
Check: 2(30) + 60 = 120 ✓ Sanity: w = 20 → A = 1600; w = 40 → A = 1600. Both less than 1800. ✓ → General rule: for a 3-sided rectangle, optimal is always w = L/4, l = L/2.
Problem (Combinatorics — classic British maths puzzle): At a party, every person shakes hands with every other person exactly once. There are 45 handshakes in total. How many people are at the party?
1
Understand the Problem
Unknown: number of people n. Given: total handshakes = 45. Condition: each pair shakes once; no one shakes their own hand. → Try small cases: 2 people = 1 shake, 3 people = 3 shakes, 4 people = 6…
2
Devise a Plan
Each person shakes hands with (n−1) others. With n people: n(n−1) total handshakes — but each shake is counted twice (A→B and B→A). Formula: total = n(n−1)/2 Set equal to 45 and solve for n.
Check: 10 × 9 / 2 = 45 ✓ The "divide by 2" step is the insight — never double-count pairs. → n(n−1)/2 = C(n,2). This formula counts teams, matchups, connections — any "choose 2" situation.
Problem (Distance — iconic US/UK word problem): Two trains start from cities 300 miles apart, heading toward each other. Train A travels at 70 mph, Train B at 80 mph. How long before they meet? How far has each traveled?
1
Understand the Problem
Unknown: time t until they meet; distances d_A and d_B. Given: total gap = 300 mi; speeds 70 and 80 mph; moving toward each other. → Draw a number line with both trains closing in.
2
Devise a Plan
Key insight: their speeds add when moving toward each other. Combined closing speed = 70 + 80 = 150 mph. Time = distance ÷ closing speed.
3
Carry Out the Plan
t = 300 ÷ 150 = 2 hours Train A: 70 × 2 = 140 miles Train B: 80 × 2 = 160 miles
4
Look Back
Check: 140 + 160 = 300 ✓; 140/70 = 160/80 = 2 hrs ✓ Faster train covered more distance — makes sense. → "Speeds add when approaching, subtract when chasing." Works for boats in rivers, planes with wind, etc.
Problem (Arithmetic Series — Gauss's Schoolboy Trick): What is the sum of all integers from 1 to 100? Young Carl Friedrich Gauss reportedly solved this in seconds as a 9-year-old. Can you find his shortcut?
1
Understand the Problem
Unknown: S = 1 + 2 + 3 + ... + 100. Adding 100 terms directly is tedious — the challenge is to find the pattern that makes it instant.
2
Devise a Plan
Pair numbers from opposite ends: (1+100), (2+99), (3+98)... Each pair sums to 101. There are 50 such pairs. S = 50 × 101.
3
Carry Out the Plan
S = 50 × 101 = 5,050
4
Look Back
Formula: n(n+1)/2 = 100×101/2 = 5,050 ✓ Gauss's legend: his teacher set this to keep the class busy for an hour. Gauss finished in seconds by seeing the pairing structure. → The pairing strategy works for any arithmetic sequence. First + last always equals second + second-to-last = a constant.
Problem (Geometric Series — The Wheat and Chessboard): A king promises a peasant: place 1 grain of rice on square 1, 2 on square 2, 4 on square 3 — doubling each time. A chessboard has 64 squares. How many grains total does the king owe?
1
Understand the Problem
Sum = 1 + 2 + 4 + 8 + ... + 263 (64 terms). This is a geometric series with first term a=1 and ratio r=2.
2
Devise a Plan
Let S = 1 + 2 + 4 + ... + 263. Then 2S = 2 + 4 + ... + 264. Subtract: 2S − S = 264 − 1. Shortcut: the sum of all terms = 2 × (last term) − 1.
3
Carry Out the Plan
S = 264 − 1 = 18,446,744,073,709,551,615 ≈ 1.84 × 1019 grains — more than 1,000 years of global rice production.
4
Look Back
The last term alone (263) is larger than all previous terms combined. That's why doubling is so deceptive — each step overtakes everything before it. → For any geometric series with r=2: sum = 2×(last term) − 1. This is why compound interest, viral growth, and computing power all feel sudden — they're not linear, they're doubling.
Problem (Number Theory — The 1000 Locker Problem): A school has 1,000 lockers (numbered 1–1000, all closed) and 1,000 students. Student 1 opens every locker. Student 2 toggles every 2nd locker. Student 3 toggles every 3rd locker. … Student 1000 toggles locker 1000. Which lockers are open at the end?
1
Understand the Problem
Locker n is toggled by student k exactly when k divides n evenly. So locker n is toggled once for each of its divisors. It ends open if and only if it is toggled an odd number of times.
2
Devise a Plan
Most integers have divisors in pairs (e.g., 12: 1×12, 2×6, 3×4 → 6 divisors, even → locker 12 ends closed). Exception: perfect squares have one divisor that pairs with itself (e.g., 36: 6×6 → 6 appears once → odd total → locker 36 stays open).
3
Carry Out the Plan
Lockers that end open = perfect squares ≤ 1000: 1, 4, 9, 16, 25, 36 … 961 = 31² Total open: 31 lockers
4
Look Back
The answer has nothing to do with 1,000 — and everything to do with the structure of divisors. Key question: "When does a number have an odd number of divisors?" Only perfect squares — because √n × √n counts once, not twice. → Locker problems appear in dozens of variants. The core insight is always: pair up the divisors and find what breaks the pairing.
Problem (State Search — The Water Jug Problem): You have a 3-liter jug and a 5-liter jug, both unmarked. You have unlimited water. Using only these two jugs, measure out exactly 4 liters. (This puzzle famously appears in the film Die Hard With a Vengeance.)
1
Understand the Problem
Allowed moves: fill a jug completely; empty a jug completely; pour from one to the other (stop when source empties or target fills). State = (amount in 3L jug, amount in 5L jug). Start: (0,0). Goal: 4L in either jug.
2
Devise a Plan
Work backwards: 4 = 5 − 1. If the 5L jug is full and has 1L poured out, 4L remains. How to get exactly 1L in the 3L jug? Fill 3L → pour into 5L (which needs only 2 more) → 1L left in 3L. Empty 5L. Pour the 1L in. Now fill 3L and pour into 5L → 4L.
3
Carry Out the Plan
Fill 3L → (3,0) → pour into 5L → (0,3) → fill 3L → (3,3) → pour 2L into 5L → (1,5) → empty 5L → (1,0) → pour 1L into 5L → (0,1) → fill 3L → (3,1) → pour into 5L → (0,4) ✓
4
Look Back
Why is 4L reachable? GCD(3, 5) = 1, which divides 4. In general: you can measure x liters with an a-liter and b-liter jug iff GCD(a, b) divides x. → "Work backwards from the goal" is one of Polya's most powerful strategies. Here: 4 = 5−1 → get 1L in 3L jug → the rest falls into place.
Problem (Probability — The Birthday Paradox): In a room of 23 people, what is the probability that at least two share the same birthday? Assume 365 days per year. Most people guess around 5–10% — the real answer will surprise you.
1
Understand the Problem
P(at least 2 share a birthday) is hard to compute directly. Use the complement: P(at least one match) = 1 − P(all birthdays are different).
2
Devise a Plan
P(all different) = (365/365) × (364/365) × (363/365) × ... × (343/365) Person 1 can be any day; person 2 must avoid 1 day; person 3 must avoid 2 days; and so on for 23 people.
23 people → over 50% chance! With 50 people → 97%. Why so counterintuitive? 23 people create C(23,2) = 253 pairs. Each pair has ~1/365 chance. 253 chances compound quickly. → "1 − P(complement)" is the universal tool for "at least one" probability problems. Always try the complement first when direct calculation is messy.
📜 Sun Tzu's Mathematical Classic (孙子算经), ~300 CE — one of China's oldest surviving math texts
Problem (鸡兔同笼 — Chickens and Rabbits in the Same Cage): A cage contains chickens and rabbits. Counting heads from above: 35 heads. Counting legs from below: 94 legs. How many chickens and how many rabbits?
1
Understand the Problem
Chickens: 2 legs. Rabbits: 4 legs. Total animals (heads): 35. Total legs: 94. Unknown: c = chickens, r = rabbits.
2
Devise a Plan
Ancient Chinese method — "make all animals stand on two legs": If all 35 are chickens: 35×2 = 70 legs. Actual = 94, gap = 94−70 = 24 "extra" legs. Each rabbit contributes 2 extra legs (4 instead of 2). Rabbits = 24 ÷ 2 = 12.
This is structurally identical to Problem 1 (Coins) — a "two types, two constraints" mixture, independently invented 1,700 years apart on opposite sides of the world. The "assume all one type, adjust for excess" strategy is universal. → The algebraic approach: c+r=35 and 2c+4r=94. The Chinese method is equivalent but requires no variables — it reasons geometrically with "excess legs."
📜 Zhang Qiu-jian's Mathematical Manual (张丘建算经), 475 CE — the first ancient problem known to have multiple valid solutions
Problem (百鸡问题 — The Hundred Fowls Problem): Roosters cost 5 coins each. Hens cost 3 coins each. Chicks cost 1 coin for 3. Buy exactly 100 birds for exactly 100 coins. Find all possible combinations of roosters, hens, and chicks.
1
Understand the Problem
r = roosters, h = hens, c = chicks. Two constraints: r + h + c = 100 (total birds) 5r + 3h + c/3 = 100 (total cost) All must be non-negative integers; c divisible by 3.
2
Devise a Plan
Multiply cost equation by 3 → 15r + 9h + c = 300. Subtract bird equation: 14r + 8h = 200 → 7r + 4h = 100. This Diophantine equation has integer solutions only when (100−7r) is divisible by 4. Since 7r ≡ 3r (mod 4), need 3r ≡ 0 (mod 4) → r must be a multiple of 4.
Zhang Qiu-jian was the first mathematician to systematically find and record all solutions — a radical idea when problems were expected to have exactly one answer. This is the earliest known Diophantine system documented with multiple solutions. → Pattern: valid values of r step by 4. Why? Because adding 4 roosters costs 28 extra coins and needs 4 fewer birds — exactly balanced by removing 4 hens×3 + adding chicks.
📜 Sun Tzu's Mathematical Classic (孙子算经), ~300 CE — origin of the Chinese Remainder Theorem, now used in RSA encryption and distributed computing
Problem (孙子算经 — Sun Tzu's Mystery Number): I have a number. Divided by 3, remainder 2. Divided by 5, remainder 3. Divided by 7, remainder 2. What is the smallest positive number satisfying all three?
1
Understand the Problem
Find the smallest n > 0 satisfying: n ≡ 2 (mod 3), n ≡ 3 (mod 5), n ≡ 2 (mod 7). This is a system of simultaneous congruences.
2
Devise a Plan
Start with the largest modulus: n ≡ 2 (mod 7) means n = 7k + 2 for integer k ≥ 0. Try k = 0, 1, 2, 3... until n ≡ 3 (mod 5). Then check mod 3.
3
Carry Out the Plan
k=0: n=2, mod 5 = 2 ✗ k=1: n=9, mod 5 = 4 ✗ k=2: n=16, mod 5 = 1 ✗ k=3: n=23, mod 5 = 3 ✓ Check mod 3: 23 = 7×3 + 2, remainder = 2 ✓ Answer: n = 23
4
Look Back
All solutions: 23, 128, 233... (repeating every 3×5×7 = 105). Sun Tzu's formula: "70×2 + 21×3 + 15×2 = 233 ≡ 23 (mod 105)" — this was formalized 1,500 years later by Gauss as the Chinese Remainder Theorem. → Today the CRT powers RSA encryption, fast multiplication of large numbers, and distributed hash tables. A 1,700-year-old puzzle underpins the internet.
📜 The Lo Shu (洛书, "River Luo Writing"), legendary origin ~2000 BCE — the world's oldest known magic square, said to have appeared on a tortoise shell
Problem (洛书 — The Lo Shu Magic Square): Arrange the numbers 1 through 9 in a 3×3 grid so that every row, every column, and both main diagonals all add up to the same sum. What is that sum, and what arrangement works?
1
Understand the Problem
9 distinct numbers in a 3×3 grid. All 8 "lines" (3 rows + 3 columns + 2 diagonals) must sum to the same value S. First determine S, then find a valid arrangement.
2
Devise a Plan
Total = 1+2+...+9 = 45. Three rows cover all 9 cells → S = 45/3 = 15. The center cell appears in 4 of the 8 lines (1 row, 1 col, 2 diagonals) — more than any other cell. Which number must go there?
3
Carry Out the Plan
Sum of all 8 lines = 8×15 = 120. Each corner contributes to 3 lines, each edge to 2, center to 4. Algebra shows center = 5. Opposite pairs always sum to 10: (1,9),(2,8),(3,7),(4,6). Solution: 2 9 4 / 7 5 3 / 6 1 8 (all rows, cols, diagonals = 15 ✓)
4
Look Back
All 8 valid arrangements are rotations/reflections of this single pattern. Even numbers (2,4,6,8) always occupy corners; odd non-center numbers (1,3,7,9) always occupy edges. The Lo Shu has been studied across cultures for 4,000 years — influencing Islamic mathematics, European Renaissance art, and the game of tic-tac-toe. → The center-must-be-5 proof: 5 is the only number that appears in 4 "lines." The average of all 9 numbers is 5 — and 5 must be in the "average position."
📜 Nine Chapters on the Mathematical Art (九章算术), ~200 BCE — Chapter 7: 盈不足 (Surplus and Deficit). Ancestor of linear algebra.
Problem (盈不足 — The Surplus-Deficit Method): Students try to share exercise books equally. If each takes 8 books, there are 3 books left over. If each takes 9 books, they are 4 books short. How many students, and how many books?
1
Understand the Problem
s = students, b = books. 8s + 3 = b (surplus of 3 books) 9s − 4 = b (deficit of 4 books) Both expressions equal b.
2
Devise a Plan
The 盈不足 formula from Nine Chapters: Students = (surplus + deficit) ÷ (difference per student) = (3 + 4) ÷ (9 − 8) Total imbalance divided by how much each student changes the balance.
3
Carry Out the Plan
s = 7 ÷ 1 = 7 students; b = 8×7 + 3 = 59 books. Check: 9×7 = 63; 63 − 59 = 4 short ✓
4
Look Back
Nine Chapters generalized this formula for any "two-scenario surplus/deficit" problem. It is the ancestor of solving 8s+3 = 9s−4 algebraically — without variables, just reasoning about imbalance. This predates Al-Khwarizmi's Algebra (820 CE) by ~1,000 years. → Formula: n = (surplus + deficit) / (rate difference). Works whenever two different rates give one too many and one too few.
📜 Nine Chapters on the Mathematical Art (九章算术), ~200 BCE — a classic "two conditions, two unknowns" problem solved by elimination
Problem (绳量井深 — The Well and the Rope): A rope measures the depth of a well. Folded in half and dangled, it extends 3 feet below the well's edge. Folded in thirds, it extends 1 foot below the edge. What is the rope's length? What is the well's depth?
1
Understand the Problem
L = rope length, D = well depth (both in feet). Folded in half: L/2 = D + 3 Folded in thirds: L/3 = D + 1 Two equations, two unknowns.
2
Devise a Plan
Subtract the two equations to eliminate D — the depths cancel: (L/2) − (L/3) = (D+3) − (D+1) = 2 L × (1/2 − 1/3) = 2 → L/6 = 2 → solve for L.
3
Carry Out the Plan
L = 12 feet. D = L/2 − 3 = 6 − 3 = 3 feet. Check: L/3 = 4; D + 1 = 4. ✓
4
Look Back
"Subtract to eliminate" is the core technique of simultaneous equations — the same operation used in Gaussian elimination (see Problem 20) and every linear algebra textbook. Nine Chapters used it ~200 BCE, centuries before the technique appeared in European mathematics. → When two equations share an unwanted variable, subtract them. The shared variable vanishes, leaving only what you need.
📜 Nine Chapters on the Mathematical Art (九章算术), ~200 BCE — China's oldest recorded pursuit problem
Problem (追及问题 — Hound and Hare Pursuit): A hare runs at 50 paces per minute with a 100-pace head start. A hound chases at 125 paces per minute. How many minutes until the hound catches the hare? How far has the hound run?
1
Understand the Problem
Both move in the same direction. Initial gap: 100 paces. Hound gains on the hare at 125 − 50 = 75 paces per minute (the closing rate).
2
Devise a Plan
Key insight: for same-direction pursuit, effective closing speed = faster speed − slower speed. Time to close gap = gap ÷ closing rate = 100 ÷ 75.
3
Carry Out the Plan
t = 100/75 = 4/3 minutes (≈ 1 min 20 sec) Hound runs: 125 × 4/3 = 166⅔ paces Verify: hare's final position = 100 (head start) + 50×4/3 = 166⅔ paces. Same position. ✓
4
Look Back
Compare with Problem 5 (Two Trains): when moving toward each other, speeds add; chasing in the same direction, speeds subtract. The same formula, discovered independently across 2,000 years and multiple civilizations. → Closing rate = |v₁ − v₂| (same direction) or v₁ + v₂ (opposite). Time = gap ÷ closing rate. Universal for any pursuit or meet problem.
📜 Legend of General Han Xin (韩信点兵), Han Dynasty ~206 BCE — a Chinese Remainder Theorem problem with a beautiful hidden shortcut
Problem (韩信点兵 — Han Xin's Troop Count): General Han Xin counted his soldiers in rows. Rows of 3: remainder 2. Rows of 5: remainder 4. Rows of 7: remainder 6. Rows of 11: remainder 10. The army has fewer than 2,000 soldiers. How many?
1
Understand the Problem
Find n < 2000 where n≡2(mod 3), n≡4(mod 5), n≡6(mod 7), n≡10(mod 11). Look at the remainders carefully: each one equals its modulus minus 1. That means n+1 ≡ 0 (mod 3), n+1 ≡ 0 (mod 5), n+1 ≡ 0 (mod 7), n+1 ≡ 0 (mod 11).
2
Devise a Plan
Elegant shortcut: n+1 must be divisible by 3, 5, 7, AND 11 simultaneously. So n+1 = LCM(3, 5, 7, 11) × k for some positive integer k.
3
Carry Out the Plan
LCM(3,5,7,11) = 3×5×7×11 = 1,155 (all prime, so LCM = product). k=1: n = 1155−1 = 1,154. k=2: n = 2309 > 2000. ✗ Answer: 1,154 soldiers.
4
Look Back
Verify: 1154÷3=384r2✓ 1154÷5=230r4✓ 1154÷7=164r6✓ 1154÷11=104r10✓ The key Polya insight: notice that all remainders = modulus−1. That observation reduces a hard CRT problem to a trivial LCM calculation. → Always examine the structure of the remainders before applying brute-force CRT. Patterns in the data often reveal elegant shortcuts.
📜 Liu Hui (刘徽), 263 CE — China's greatest classical mathematician. His 割圆术 (circle-cutting method) computed π to 5 decimal places, 1,200 years before European mathematicians reached the same precision.
Problem (割圆术 — Liu Hui's Circle-Cutting Method): A regular hexagon inscribed in a unit circle (radius = 1) has perimeter = 6. Liu Hui doubled the sides repeatedly: 12-gon, 24-gon, 48-gon, 96-gon. His 96-sided polygon had perimeter ≈ 3.14103. Why does this approximate π, and how does it work?
1
Understand the Problem
The circumference of a unit circle = 2π. An inscribed polygon's perimeter is always less than the circumference, but approaches it as the number of sides increases. So π ≈ (polygon perimeter) / 2 for a large-sided polygon in a unit circle.
2
Devise a Plan
Start with the hexagon (side = radius = 1, perimeter = 6). To double the sides: use the Pythagorean theorem on the triangle formed by the old side and the new shorter sides (no trigonometry needed — Liu Hui only had geometry). Each doubling: new side² = r² − (r − h)², where h is the height from center to old side.
As n→∞, the polygon's perimeter → 2π (since n·sin(π/n) → π). Liu Hui also showed that a circumscribed polygon gives an upper bound — trapping π between two closing values. This "squeezing" idea is the basis of the squeeze theorem in calculus. → Computing π to arbitrary precision by polygon doubling was the state of the art for ~1,400 years. Liu Hui's method predates Archimedes' known Chinese translation by centuries.
📜 Nine Chapters on the Mathematical Art (九章算术), ~200 BCE — Chapter 8 (方程, "Rectangular Arrays") contains the world's first recorded use of Gaussian elimination, ~2,000 years before Gauss (1809 CE)
Problem (方程 — Three Grades of Grain): Three grain harvests are mixed in different ratios: 3 loads top-grade + 2 medium + 1 low = 39 shi 2 top + 3 medium + 1 low = 34 shi 1 top + 2 medium + 3 low = 26 shi How much does each grade yield per load?
1
Understand the Problem
t = top, m = medium, l = low (yield in shi per load). 3t + 2m + l = 39 …(1) 2t + 3m + l = 34 …(2) t + 2m + 3l = 26 …(3)
2
Devise a Plan
Nine Chapters used 方程术: write coefficients in a rectangular column array and eliminate variables one at a time — exactly what we now call Gaussian elimination, named after Gauss (1809 CE), 2,000 years later. Eliminate t first: multiply equation (3) by 2, subtract (2); multiply (3) by 3, subtract (1).
3
Carry Out the Plan
2×(3) − (2): m + 5l = 18 …(4) 3×(3) − (1): 4m + 8l = 39 …(5) 4×(4) − (5): 12l = 33 → l = 2¾ From (4): m = 18 − 5(11/4) = 4¼ From (3): t = 26 − 2(17/4) − 3(11/4) = 9¼
4
Look Back
Check (1): 3(9.25)+2(4.25)+2.75 = 27.75+8.5+2.75 = 39 ✓ Nine Chapters wrote coefficients in what was effectively the world's first matrix and performed column operations. The method is equivalent to the augmented-matrix approach taught in every linear algebra course today — first documented ~200 BCE. → This 3×3 technique scales to n×n. Every engineering, economics, and physics model with multiple unknowns ultimately uses Gaussian elimination.
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❓Frequently Asked Questions
What is Polya's problem-solving method?
George Polya's method, introduced in his 1945 book How to Solve It, breaks any problem into four steps: (1) Understand the Problem, (2) Devise a Plan, (3) Carry Out the Plan, and (4) Look Back. It works for any discipline — not just math — and emphasizes thinking process over memorizing procedures. The book has sold over one million copies and been translated into 20+ languages.
What are the 4 steps of Polya's framework?
Step 1 — Understand the Problem: What is the unknown? What is given? What is the condition? Can you draw a figure or restate it in your own words? Step 2 — Devise a Plan: Have you seen a similar problem? Can you solve a simpler version first? Work backwards? Use an auxiliary element? Step 3 — Carry Out the Plan: Execute step by step, justifying each action. Where exactly does it break down? Step 4 — Look Back: Verify the result. Can you solve it another way? Can the method generalize? Add it to your mental toolkit.
What are classic Chinese math problems in English?
This tool features 10 ancient Chinese math problems translated and solved in English: Chickens & Rabbits (鸡兔同笼, ~300 CE), Hundred Fowls (百鸡问题, 475 CE), Sun Tzu's Mystery Number (origin of the Chinese Remainder Theorem), Lo Shu Magic Square (洛书, ~2000 BCE), the Surplus-Deficit method from Nine Chapters (盈不足, ~200 BCE), the Well and the Rope, Hound and Hare pursuit, Han Xin's Troop Count (韩信点兵), Liu Hui's π approximation (割圆术, 263 CE), and Three Grades of Grain — the world's first Gaussian elimination (~200 BCE).
What is the Chickens and Rabbits problem (鸡兔同笼)?
The Chickens and Rabbits problem (鸡兔同笼, literally "chickens and rabbits in the same cage") is from Sun Tzu's Mathematical Classic (~300 CE), one of China's oldest math texts. A cage has chickens (2 legs) and rabbits (4 legs). Given the total number of heads and legs, find how many of each there are. The ancient Chinese solution — assume all are one type, then adjust for the excess legs — is identical in structure to classic Western mixture problems, independently discovered 1,700 years later.
What is the Chinese Remainder Theorem?
The Chinese Remainder Theorem (CRT) originates from Sun Tzu's Mathematical Classic (~300 CE): given a number that leaves specific remainders when divided by 3, 5, and 7, find the number. Sun Tzu developed a method to solve this class of problem. The theorem was formalized by Gauss in 1801 and named in honor of its Chinese origin. Today it is a cornerstone of RSA encryption, fast large-number multiplication, distributed computing, and error-correcting codes.
What is the Lo Shu Magic Square (洛书)?
The Lo Shu (洛书, "River Luo Writing") is the world's oldest known magic square: a 3×3 grid of digits 1–9 where every row, column, and diagonal sums to 15. According to Chinese legend it appeared on a tortoise shell around 2000 BCE. The unique arrangement (2-9-4 / 7-5-3 / 6-1-8) has only 8 valid forms (rotations and reflections). It influenced Chinese philosophy, feng shui, Islamic mathematics, and European Renaissance art.
How do you solve the Locker Problem?
In the classic 1,000-locker problem, locker n ends open if and only if it has an odd number of divisors. Most integers have divisors in pairs (e.g., 12: 1×12, 2×6, 3×4 → 6 divisors, even → closed). Perfect squares are the exception — the square root pairs with itself, giving an odd count (e.g., 36: 1,2,3,4,6,9,12,18,36 → 9 divisors → open). So the open lockers are exactly the perfect squares: 1, 4, 9, ..., 961 = 31² — a total of 31 lockers.
Why does the Birthday Paradox work with only 23 people?
With 23 people, there are C(23,2) = 253 distinct pairs, and each pair has roughly a 1/365 chance of sharing a birthday. The probability that all 253 pairs have different birthdays is only about 49.3%, so the chance of at least one shared birthday exceeds 50%. The paradox lies in pair-counting: 23 people create far more chances than intuition suggests. With 50 people the probability rises to 97%.
What were Nine Chapters on the Mathematical Art?
Nine Chapters on the Mathematical Art (九章算术) is China's most important classical mathematics text, compiled around 200 BCE. It covers proportions, geometry, systems of equations, and much more. Key achievements: the Surplus-Deficit method (ancestor of algebra), the world's first matrix-based Gaussian elimination (Chapter 8, 方程), and systematic use of fractions and negative numbers. It served as the standard Chinese mathematics textbook for over 1,000 years.
What makes this AI math tutor different from ChatGPT or Wolfram Alpha?
Most AI tools solve problems for you. This tutor uses Polya's method to guide your thinking: it asks questions instead of giving answers, so you do the reasoning and arrive at the insight yourself. Research consistently shows that productive struggle — working through difficulty — builds mathematical thinking far better than copying solutions. The AI acts like a patient senior student who asks "What do you know so far?" rather than doing your homework.
Who made this?
I'm an ordinary person who finds math more interesting than the way it's usually taught. This tool is my attempt to make Polya's 4-step method — one of the most useful thinking frameworks ever written — actually interactive. Built on Polya's How to Solve It (1945).