Intermediate algebra problems (ages 13 to 16)

Answer: D, 119

1. Square x − 1/x = 3: x² − 2 + 1/x² = 9.
2. So x² + 1/x² = 11.
3. Square again: x⁴ + 2 + 1/x⁴ = 121.
4. So x⁴ + 1/x⁴ = 119 (D).

Why this works: Squaring x ± 1/x always produces a constant cross term (±2), so you can climb to higher powers without ever finding x.

Where it leads: Repeated squaring like this gives x^2k + x^−2k quickly; a recurrence handles all powers.

Answer: 10

1. For x² − 7x + k = 0 the roots add to 7 and multiply to k.
2. The roots add to 7 and differ by 3, so the larger is (7 + 3)/2 = 5 and the smaller is (7 − 3)/2 = 2.
3. k = 5 × 2 = 10.
4. Check: x² − 7x + 10 = (x − 2)(x − 5). ✓ Answer 10.

Why this works: For x² − sx + p = 0 the roots add to s and multiply to p. Using these facts is often faster than the quadratic formula.

Where it leads: Vieta’s formulas: for x² − sx + p, the roots add to s and multiply to p. (Difference)² = s² − 4p.

Answer: C, 8

1. f(f(x)) = a(ax + b) + b = a²x + ab + b.
2. This equals 9x + 8 for every x, so the coefficients match: a² = 9 and (a + 1)b = 8.
3. a > 0, so a = 3, and then 4b = 8, b = 2.
4. So f(x) = 3x + 2. Check: f(f(x)) = 3(3x + 2) + 2 = 9x + 8. ✓
5. f(2) = 6 + 2 = 8 (C).

Why this works: Two polynomials that agree for every x have the same coefficients, so one identity gives several equations.

Where it leads: With a < 0 there is another solution, f(x) = −3x − 4. Functional square roots of linear maps come in pairs.

Answer: B, x = 9

1. Square both sides: x + 7 = (x − 5)² = x² − 10x + 25.
2. Rearrange: x² − 11x + 18 = 0, i.e. (x − 2)(x − 9) = 0, so x = 2 or x = 9.
3. Squaring can add false solutions, so check both in the original equation.
4. x = 9: √16 = 4 and 9 − 5 = 4. ✓
5. x = 2: √9 = 3 but 2 − 5 = −3. ✗ (A square root is never negative.)
6. Only x = 9 (B).

Why this works: Squaring can create false solutions, because a = b and a = −b square to the same thing. A square root is never negative, so the right side must be ≥ 0.

Where it leads: Squaring can introduce false solutions, because √ is never negative; always check in the original equation.

Answer: 550

1. Let the first term be a and the common difference d. Sum of the first n terms: S_n = (n/2)(2a + (n − 1)d).
2. Terms 11–20 add to S₂₀ − S₁₀ = 500 − 150 = 350.
3. Each term in a block of ten is 10d bigger than the matching term ten places earlier, so each block’s sum is 100d bigger than the one before.
4. 350 − 150 = 100d, so 100d = 200 (d = 2, and from S₁₀ = 150, a = 6).
5. Terms 21–30: 350 + 200 = 550. (Check: S₃₀ = 15(12 + 58) = 1050 = 500 + 550.)

Why this works: Consecutive equal-length blocks of an arithmetic sequence form another arithmetic sequence. Seeing the structure saves solving for the first term at all.

Where it leads: Sums of consecutive blocks of an arithmetic sequence form another arithmetic sequence, with common difference (block length)² × d.

Answer: D, 24

1. Let u = 2x and v = 3y. Then u + v = 24 and xy = uv/6.
2. For a fixed sum, a product is largest when the two parts are equal (AM–GM: uv ≤ ((u + v)/2)² = 144).
3. So uv ≤ 144, with equality when u = v = 12.
4. Then x = 6, y = 4, and xy = 144/6 = 24.
5. The largest value of xy is 24 (D).

Why this works: For a fixed sum, a product is largest when the parts are equal. Choosing parts (2x and 3y) whose sum is fixed makes the idea apply.

Where it leads: For a fixed sum, a product is largest when the parts are equal: the AM–GM inequality.

Answer: 10

1. |x − 3| + |x + 2| is the distance from x to 3 plus the distance from x to −2 on the number line.
2. For −2 ≤ x ≤ 3 the two distances add to exactly 5, which is less than 11: x = −2, −1, 0, 1, 2, 3 all work (6 numbers).
3. For x > 3: (x − 3) + (x + 2) = 2x − 1 < 11 gives x < 6: x = 4, 5.
4. For x < −2: (3 − x) + (−2 − x) = 1 − 2x < 11 gives x > −5: x = −4, −3.
5. Total: 6 + 2 + 2 = 10 whole numbers (−4 to 5).

Why this works: Reading |x − a| as a distance turns the inequality into a picture: the sum of distances to two points is constant between them and grows by 2 per step outside.

Where it leads: Thinking of |x − a| as distance on a number line turns absolute-value inequalities into pictures.

Answer: 133

1. (x + y)² = x² + y² + 2xy, so 49 = 29 + 2xy and xy = 10.
2. (x + y)³ = x³ + y³ + 3xy(x + y), so x³ + y³ = 343 − 3 × 10 × 7.
3. = 343 − 210 = 133. (Check: x, y = 2, 5 gives 8 + 125 = 133.)

Why this works: Symmetric expressions can be built from the sum and the product, so you never need x and y individually.

Where it leads: Newton’s identities give xⁿ + yⁿ from x + y and xy for every n, through the recurrence p_n = (x + y)p_n−1 − xy p_n−2.

Answer: x = 5

1. 4^{x − 1} = (2²)^{x − 1} = 2^{2x − 2}.
2. So 2^{x + 3} = 2^{2x − 2}, and the powers must be equal: x + 3 = 2x − 2.
3. x = 5. Check: 2⁸ = 256 = 4⁴. ✓

Why this works: Writing both sides with the same base turns an equation in powers into an equation in exponents.

Where it leads: When the bases cannot be matched (2^x = 3), logarithms do the same job: x = log 3 / log 2.

Answer: 1272

1. Number of terms: (99 − 7)/4 + 1 = 23 + 1 = 24.
2. Pair the first and last, second and second-last, …: each pair sums to 7 + 99 = 106, and there are 12 pairs.
3. Sum = 12 × 106 = 1272.

Why this works: An arithmetic sequence pairs up from the two ends into equal sums, giving (number of terms) × (first + last)/2.

Where it leads: The ‘+ 1’ when counting terms is the fence-post rule again. Sums of arithmetic sequences also give the area of a trapezium.

Answer: C, 3

1. From the 3rd term to the 6th you multiply by r three times: r³ = 96/12 = 8, so r = 2 (the only real cube root).
2. The 3rd term is ar² = 4a = 12, so a = 3.
3. The first term is 3 (C): 3, 6, 12, 24, 48, 96.

Why this works: Dividing two terms cancels the first term and leaves a power of the ratio.

Where it leads: With complex numbers, r³ = 8 has two more solutions, 2ω and 2ω², giving complex geometric sequences with the same 3rd and 6th terms.

Answer: 8

1. Two different real roots exactly when the discriminant k² − 4 × 9 is positive: k² > 36.
2. So |k| > 6: k = 7, 8, 9, 10 or −7, −8, −9, −10.
3. That is 8 values.

Why this works: The discriminant b² − 4ac decides the number of real roots without solving the equation.

Where it leads: At k = ±6 the roots coincide (x = ∓3): the parabola just touches the x-axis. Tangency problems often reduce to ‘discriminant = 0’.

Answer: 99/100

1. Factorise: (x − 3)(x + 3) / ((x + 3)(x − 2)) = (x − 3)/(x − 2), for x ≠ −3.
2. At x = 102: 99/100.
3. The value is 99/100.

Why this works: Cancelling a common factor before substituting avoids big arithmetic.

Where it leads: The simplified form (x − 3)/(x − 2) is not defined at x = 2, and the original is also undefined at x = −3: cancelling can hide a ‘hole’ in a graph.

Answer: B, −2

1. f(g(x)) = 2x² + 1 and g(f(x)) = (2x + 1)² = 4x² + 4x + 1.
2. Setting them equal: 2x² + 4x = 0, so 2x(x + 2) = 0, giving x = 0 or x = −2.
3. The sum is −2 (B).

Why this works: Composite functions are applied inside-out; once written out, the equation is an ordinary quadratic.

Where it leads: f(g(x)) and g(f(x)) are usually different: composition is not commutative. Pairs of functions that do commute are rare and interesting.

Answer: 16

1. (√5 + √3)² = 5 + 2√15 + 3 = 8 + 2√15.
2. (√5 − √3)² = 8 − 2√15.
3. Sum: 16.

Why this works: (a + b)² + (a − b)² = 2(a² + b²): the cross terms always cancel.

Where it leads: The same identity is the parallelogram law: the squares of the diagonals of a parallelogram add up to the sum of the squares of its four sides.

Answer: 2

1. Replace 5 by 3^a in the second equation: (3^a)^b = 9.
2. So 3^ab = 3².
3. ab = 2.

Why this works: Substituting one exponential into another multiplies the exponents, which is exactly the product we want.

Where it leads: In log form, a = log₃ 5 and b = log₅ 9, and ab = log₃ 9 = 2: the change-of-base rule log₃ 5 × log₅ 9 = log₃ 9.

Answer: B, 11/20

1. 1 − 1/n² = (n − 1)(n + 1)/n² = (n − 1)/n × (n + 1)/n.
2. The product of the (n − 1)/n factors from n = 2 to 10 telescopes to 1/10. The product of the (n + 1)/n factors telescopes to 11/2.
3. Total: 1/10 × 11/2 = 11/20 (B).

Why this works: Factorising each term into two fractions makes both products telescope.

Where it leads: Continuing to infinity, the product tends to 1/2. Infinite products like this (and Euler’s product for sin x) are a powerful tool in analysis.

Answer: 5

1. 2x − 5 = 7 gives x = 6.
2. 2x − 5 = −7 gives x = −1.
3. The sum is 6 + (−1) = 5.

Why this works: Absolute value measures distance, so |2x − 5| = 7 has a solution on each side of x = 2.5; they are symmetric about 2.5, so they add to 5.

Where it leads: Thinking of |x − a| as the distance from a solves harder problems quickly, like minimising |x − 1| + |x − 4| + |x − 9| (answer: at the median, x = 4).

Answer: 7

1. x² − y² = (x − y)(x + y) = 40, so 4(x + y) = 40 and x + y = 10.
2. Adding x + y = 10 and x − y = 4: 2x = 14.
3. x = 7 (and y = 3).

Why this works: The difference of two squares links the given difference to the sum, turning a quadratic system into a linear one.

Where it leads: This is how Fermat factorised numbers: write n = x² − y² = (x − y)(x + y), searching x upwards from √n.

Answer: C, 5

1. f(5) = 1/(1 − 5) = −1/4.
2. f(−1/4) = 1/(1 + 1/4) = 4/5.
3. f(4/5) = 1/(1 − 4/5) = 5 (C).

Why this works: Applying f three times returns to the start. In fact f(f(f(x))) = x for every x where it is defined.

Where it leads: Functions with f⁽ⁿ⁾ = identity have finite order. 1/(1 − x) permutes the three points 0, 1 and ∞ of the projective line.

Answer: 6

1. The product is negative when exactly one factor is negative.
2. x − 2 < 0 and x − 9 > 0 is impossible; so x − 2 > 0 and x − 9 < 0: 2 < x < 9.
3. Whole numbers: 3, 4, 5, 6, 7, 8: 6.

Why this works: The sign of a product changes only where a factor is zero, so the roots split the number line into intervals of constant sign.

Where it leads: A sign table at the roots solves any polynomial inequality. For a quadratic with positive x² coefficient, ‘< 0’ is always the interval between the roots.

Answer: 2047

1. Call the sum S. Then S + 1 = 1 + 1 + 2 + 4 + … + 2¹⁰.
2. The first two terms make 2, which with the next makes 4, then 8, …, finally 2¹⁰ + 2¹⁰ = 2¹¹.
3. So S = 2¹¹ − 1 = 2047.

Why this works: Adding 1 makes the terms snowball, each doubling the last: the sum of powers of 2 is one less than the next power.

Where it leads: In binary, 2047 is 11111111111 (eleven 1s), and adding 1 carries all the way to 100000000000. The general formula is a(rⁿ − 1)/(r − 1).

Answer: B, 2 hours

1. A fills 1/3 of the tank per hour; B fills 1/6.
2. Together: 1/3 + 1/6 = 1/2 of the tank per hour.
3. So the tank takes 2 hours (B).

Why this works: Rates (tanks per hour) add; times do not. Convert to rates, add, then convert back.

Where it leads: The combined time is 1/(1/a + 1/b) = ab/(a + b), half the harmonic mean, the same algebra as resistors in parallel.

Answer: 7/10

1. b = (2/5)c, and a = (3/4)b = (3/4)(2/5)c = (3/10)c.
2. a + b = (3/10)c + (4/10)c = (7/10)c.
3. So (a + b)/c = 7/10.

Why this works: Expressing everything in terms of one quantity lets the ratios combine.

Where it leads: Chained ratios are the same as multiplying fractions: a/c = (a/b)(b/c). This is the idea behind units cancelling in physics.

Answer: 4

1. Let u = x²: u² − 5u + 4 = 0, so (u − 1)(u − 4) = 0 and u = 1 or 4.
2. x² = 1 gives x = ±1; x² = 4 gives x = ±2.
3. So there are 4 real solutions.

Why this works: A substitution reveals a hidden quadratic; each positive value of u gives two values of x.

Where it leads: If one value of u had been negative, it would give no real x. Counting real roots of equations like this is the first step towards Descartes’ rule of signs.

Answer: D, 4051

1. 2026² − 2025² = (2026 − 2025)(2026 + 2025).
2. = 1 × 4051.
3. = 4051 (D).

Why this works: Factorising first replaces two large squarings with one easy multiplication.

Where it leads: Consecutive squares differ by consecutive odd numbers, which is why 1 + 3 + 5 + … + (2n − 1) = n².

Answer: 515

1. Let v_n = u_n − 3. Then v_n+1 = u_n+1 − 3 = 2u_n − 6 = 2v_n.
2. v₁ = 1, so v_n = 2ⁿ⁻¹ and v₁₀ = 512.
3. u₁₀ = 512 + 3 = 515.

Why this works: Shifting by the fixed point (3, where u = 2u − 3) turns the recurrence into a plain geometric sequence.

Where it leads: Every recurrence u_n+1 = au_n + b with a ≠ 1 can be solved this way, shifting by b/(1 − a). It models loans, cooling and populations.

Answer: 1/2

1. Gradient = (9 − 3)/(5 − 2) = 2.
2. From (2, 3), to drop by 3 to the x-axis you move 3/2 to the left.
3. p = 2 − 3/2 = 1/2. (Equation: y = 2x − 1.)

Why this works: The gradient tells you how far to move across for each unit down, so the intercept follows without writing the full equation.

Where it leads: The x-intercept of y = mx + c is −c/m. Newton’s method for solving equations repeatedly finds where a tangent line crosses the x-axis.