Stanford Math Tournament style problems across Team, General, Guts, and Power rounds. Try each problem first, then open the strategy analysis and worked solution.
Problem 1SMT · Team · Counting with stars and bars
How many 5-digit positive integers have digits that add up to 12?
Strategy analysis
THE TRICK: Shift the first digit, then use stars and bars with inclusion–exclusion. Writing the first digit as 1 + e₁ turns "5-digit number" into "five nonnegative numbers summing to 11," and the only overcount comes from digits reaching 10, which you subtract case by case.
Worked solution and answer
Let the digits be d₁d₂d₃d₄d₅ with d₁ ≥ 1, the rest ≥ 0, and d₁ + d₂ + d₃ + d₄ + d₅ = 12. Set e₁ = d₁ − 1 and eᵢ = dᵢ for i ≥ 2. Then e₁ + e₂ + e₃ + e₄ + e₅ = 11 with all eᵢ ≥ 0, which has C(11 + 5 − 1, 5 − 1) = C(15, 4) = 1365 solutions. Now remove the ones where some digit is at least 10. If d₁ ≥ 10, then e₁ ≥ 9: writing e₁ = 9 + f₁ leaves a sum of 2 over five variables, giving C(6, 4) = 15 solutions. If some dᵢ ≥ 10 for i ≥ 2, then eᵢ ≥ 10: writing eᵢ = 10 + fᵢ leaves a sum of 1 over five variables, giving C(5, 4) = 5 solutions for each of the 4 choices of i, or 20 total. No two of these violations can happen at once, since they would already force a sum above 11. So the valid count is 1365 − 15 − 20 = 1330.
Answer: 1330
Problem 2SMT · Team · Circle geometry and lens area
Two circles, each of radius 6, have centers 6 units apart. Find the area of the region common to both circles.
Strategy analysis
THE TRICK: Split the lens into two 120° sectors minus the rhombus they share. Because the radius equals the center distance, the triangle formed by the two centers and an intersection point is equilateral, so each sector angle is 2 × 60° = 120°.
Worked solution and answer
Let the centers be O₁ and O₂ and let P and Q be the intersection points. Triangle O₁O₂P is equilateral (all sides 6), so angle PO₁Q = 2 × 60° = 120°. The lens equals sector PO₁Q plus sector PO₂Q minus the rhombus O₁PO₂Q (counted twice). Each 120° sector has area (120/360)π·6² = 12π, and the rhombus is two equilateral triangles of side 6, with area 2 × (√3/4)·6² = 18√3. So the lens area is 12π + 12π − 18√3 = 24π − 18√3.
Answer: 24π − 18√3
Problem 3SMT · General · Cyclic quadrilaterals
A quadrilateral with side lengths 5, 6, 7, and 8, in that order, is inscribed in a circle. Find its area.
Strategy analysis
THE TRICK: Brahmagupta's formula — the cyclic condition is the whole problem. For a cyclic quadrilateral, the area depends only on the four side lengths through the semiperimeter, exactly like Heron's formula for triangles, so no angles or diagonals are ever needed.
Worked solution and answer
The semiperimeter is s = (5 + 6 + 7 + 8)/2 = 13. By Brahmagupta's formula, the area of a cyclic quadrilateral is √((s − a)(s − b)(s − c)(s − d)) = √((13 − 5)(13 − 6)(13 − 7)(13 − 8)) = √(8 · 7 · 6 · 5) = √1680 = √(16 · 105) = 4√105.
Answer: 4√105
Problem 4SMT · Guts · 3D geometry
Four vertices of a cube of side length 2 are chosen so that they form a regular tetrahedron. Find the volume of this tetrahedron.
Strategy analysis
THE TRICK: Pick alternating vertices and use the determinant volume formula. Opposite corners of each face are joined, so every edge of the tetrahedron is a face diagonal of the cube — equal by symmetry — and the scalar triple product turns the volume into one determinant.
Worked solution and answer
Place the cube with vertices from (0, 0, 0) to (2, 2, 2) and take the alternating vertices A = (0, 0, 0), B = (2, 2, 0), C = (2, 0, 2), D = (0, 2, 2). Each pair differs in exactly two coordinates by 2, so every edge has length √(2² + 2²) = 2√2, confirming the tetrahedron is regular. Its volume is (1/6)|det(B − A, C − A, D − A)| = (1/6)|det of the matrix with rows (2, 2, 0), (2, 0, 2), (0, 2, 2)| = (1/6)|2·(0·2 − 2·2) − 2·(2·2 − 2·0) + 0| = (1/6)|−8 − 8| = 16/6 = 8/3.
Answer: 8/3
Problem 5SMT · General · Modular arithmetic
Find the last three digits of 7²⁰²⁵.
Strategy analysis
THE TRICK: Hunt for a small power that is 1 modulo 1000 and reduce the exponent. Computing powers of 7 mod 1000 shows that 7²⁰ ≡ 1 (mod 1000), so only the remainder of 2025 upon division by 20 matters.
Worked solution and answer
We work modulo 1000. First, 7² = 49, 7⁴ = 49² = 2401 ≡ 401, 7⁵ ≡ 401 · 7 = 2807 ≡ 807, and 7¹⁰ ≡ 807² = 651249 ≡ 249, so 7²⁰ ≡ 249² = 62001 ≡ 1 (mod 1000). Since 2025 = 20 · 101 + 5, we get 7²⁰²⁵ = (7²⁰)¹⁰¹ · 7⁵ ≡ 1¹⁰¹ · 807 = 807 (mod 1000). Hence the last three digits are 807.
Answer: 807
Problem 6SMT · Team · Diophantine equations
Find the number of pairs (x, y) of positive integers with x ≤ y satisfying 1/x + 1/y = 1/6.
Strategy analysis
THE TRICK: Simon's Favorite Factoring Trick — clear denominators and factor. The equation 1/x + 1/y = 1/6 rearranges to (x − 6)(y − 6) = 36, turning a fraction equation into a divisor-counting problem.
Worked solution and answer
Multiply by 6xy: 6y + 6x = xy, so xy − 6x − 6y = 0. Adding 36 to both sides factors it: (x − 6)(y − 6) = 36. Since x, y are positive and 1/x + 1/y = 1/6 forces x, y > 6, both factors are positive divisors of 36. With x ≤ y we need x − 6 ≤ y − 6, so x − 6 is a divisor of 36 that is at most √36 = 6: namely 1, 2, 3, 4, 6, giving (x, y) = (7, 42), (8, 24), (9, 18), (10, 15), (12, 12). That is 5 pairs.
Answer: 5
Problem 7SMT · General · Polynomial roots and power sums
Let α and β be the two roots of x² − 7x + 2 = 0. Compute α⁴ + β⁴.
Strategy analysis
THE TRICK: Never solve for the roots — run the power-sum recurrence. Each root satisfies r² = 7r − 2, so the sums Sₙ = αⁿ + βⁿ obey Sₙ = 7Sₙ₋₁ − 2Sₙ₋₂, and you climb from S₁ to S₄ in three easy steps.
Worked solution and answer
By Vieta's formulas, α + β = 7 and αβ = 2. Let Sₙ = αⁿ + βⁿ. Then S₁ = 7, and S₂ = (α + β)² − 2αβ = 49 − 4 = 45. Since each root satisfies r² = 7r − 2, multiplying by rⁿ⁻² and adding gives Sₙ = 7Sₙ₋₁ − 2Sₙ₋₂. Thus S₃ = 7·45 − 2·7 = 301, and S₄ = 7·301 − 2·45 = 2107 − 90 = 2017.
Answer: 2017
Problem 8SMT · Team · Systems and symmetric sums
Real numbers x and y satisfy x² + xy + y² = 7 and x² − xy + y² = 3. Compute x⁴ + y⁴.
Strategy analysis
THE TRICK: Add and subtract the two equations before doing anything else. Adding isolates x² + y², subtracting isolates xy — and x⁴ + y⁴ is just (x² + y²)² − 2x²y², so you never need x and y themselves.
Worked solution and answer
Adding the equations gives 2(x² + y²) = 10, so x² + y² = 5. Subtracting the second from the first gives 2xy = 4, so xy = 2. Then x⁴ + y⁴ = (x² + y²)² − 2(xy)² = 5² − 2·2² = 25 − 8 = 17.
Answer: 17
Problem 9SMT · Power · Orders modulo n
(a) Find the smallest positive integer n such that 2ⁿ ≡ 1 (mod 21). (b) Find the remainder when 2²⁰²⁵ is divided by 21.
Strategy analysis
THE TRICK: Part (a) builds the tool part (b) uses — the multiplicative order. Once you know the smallest exponent that returns 2 to 1 modulo 21, every huge power reduces to a tiny one by taking the exponent modulo that order.
Worked solution and answer
(a) Compute powers of 2 modulo 21: 2¹ = 2, 2² = 4, 2³ = 8, 2⁴ = 16, 2⁵ = 32 ≡ 11, 2⁶ = 64 ≡ 1 (mod 21). Since no smaller positive exponent gives 1, the order is n = 6. (b) Because 2⁶ ≡ 1 (mod 21), only the exponent modulo 6 matters: 2025 = 6 · 337 + 3, so 2²⁰²⁵ = (2⁶)³³⁷ · 2³ ≡ 1³³⁷ · 8 = 8 (mod 21).
Answer: (a) 6, (b) 8
Problem 10SMT · Guts · Regular polygons
A regular dodecagon (12-sided polygon) has side length 2. Find its area.
Strategy analysis
THE TRICK: Slice the dodecagon into 12 congruent isosceles triangles and rationalize through sin 15°. The central angle is 30°, and the key identity sin²15° = (2 − √3)/4 turns the circumradius into a clean surd, so the area collapses without a calculator.
Worked solution and answer
Join the center to all 12 vertices, dividing the dodecagon into 12 congruent triangles with vertex angle 360°/12 = 30°. Each triangle has two sides equal to the circumradius R and area (1/2)R²sin 30° = R²/4, so the total area is 3R². The side length satisfies 2 = 2R sin 15°, so R² = 1/sin²15°. Using sin²15° = (1 − cos 30°)/2 = (2 − √3)/4 gives R² = 4/(2 − √3) = 4(2 + √3) = 8 + 4√3. Therefore the area is 3(8 + 4√3) = 24 + 12√3.