AMC 10 · 2020 · #25

Grade 9 algebra

Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search

floor-functionquadratic-equationsvieta-formulastriangular-numbersbound-inequality-then-enumerate easier-related-problemconvert-to-algebrawork-backwards ↑ Prerequisites: floor-functionquadratic-equations

📏 Long solution 💡 4 insights

Problem

A positive fraction in lowest terms is chosen so that the real numbers x satisfying the floor of x times the fractional part of x equals that fraction times x squared add up to exactly 420. Find the sum of the fraction's numerator and denominator.

Pick an answer.

(A)

245

(B)

593

(C)

929

(D)

1331

(E)

1332

How to solve

Strategy Solve an Easier Related Problem

Tool #4 (Introduce a Variable): name the two halves of x as n = ⌊ x ⌋ and f = {x}, which turns a floor equation into ordinary algebra. Tool #3 (Eliminate Possibilities): a sign check kills every x ≤ 0, so only n ≥ 1 has to be searched. Tool #9 (Easier Problem): this is the crux — measuring the fractional part in units of n by writing f = nu makes n cancel completely, so infinitely many separate cases collapse into the single n-free equation u = a(1+u)². Tool #2 (Systematic List): the solutions are then the arithmetic list x = n(1+u) for n = 1, 2, …, N, and the given sum 420 becomes one equation in N and u. Tool #14 (Extreme Principle): N is defined by a boundary — it is the last index before the fractional part reaches 1 — and that boundary squeezes (N+1)² into a gap containing exactly one square. Tool #11 (Work Backwards): with u known, run u = a(1+u)² in reverse to recover a.

1STEP 1

Split x in two

Split into floor and fractional part.

n f = a(n+f)², 0 ≤ f ＜ 1

2STEP 2

Throw away the non-positive x

Signs leave only positives.

x ≤ 0 → nf ≤ 0 ＜ ax² (x ≠ 0) ⟹ n ≥ 1 and 0 ＜ f ＜ 1

3STEP 3

Measure in units of the floor

In units of the floor it becomes one equation.

u = a(1+u)² ⟺ a u² + (2a - 1)u + a = 0

4STEP 4

The two reciprocal roots

The roots are reciprocal, so only one is legal.

u₁ u₂ = a/a = 1 → {u₀, 1/u₀}; legality u ＜ 1 keeps exactly one

5STEP 5

List the solutions and add

The solutions line up neatly.

(1+u) · (N(N+1))/2 = 420, Nu ＜ 1 ≤ (N+1)u

6STEP 6

Squeeze with inequalities

Two inequalities pin the count exactly.

840 ＜ (N+1)² ≤ 841 → N = 28, u = 1/29

7STEP 7

Run the substitution backwards

Undoing it gives 929.

a = u/((1+u)²) = 29/900, gcd(29, 900) = 1 → p + q = 929 → (C)

Answer

929

Four checks. (1) Substitution: a = 29/900 and u = 1/29 satisfy a(1+u)² = 29/900 · 900/841 = 29/841 = 1/29 = u, so u really is a root. (2) Direct listing: the claimed solutions are x = 30n/29 for n = 1, …, 28; the largest is 840/29 ≈ 28.966, whose floor is 28 and whose fractional part is 28/29 ＜ 1, so it is legal, while n = 29 would force {x} = 29/29 = 1, which is not a fractional part — the list stops exactly where it should. (3) The sum: 30/29 · (28 · 29)/2 = 30/29 · 406 = 420 exactly, matching the condition, and adding x = 0 changes nothing. (4) Uniqueness: the squeeze 840 ＜ (N+1)² ≤ 841 admits exactly one integer, so no other a can produce the total 420 — the answer is forced, not merely consistent. A pleasant side check: a = u/((1+u)²) is unchanged when u is replaced by 1/u, so even the discarded reciprocal root points at the same a. Since p + q = 929, the answer is (C).

💡Key takeaway

Writing the fractional part as n times a ratio u makes n cancel, so one number u builds every solution x = n(1+u); the total 420 then traps (N+1)² between 840 and 841, forcing u = 1/29 and a = 29/900.

Name the two halves of x
Throw away every non-positive x
Measure f in units of n
Reciprocal roots leave exactly one
List the solutions and add them
Squeeze N between two inequalities
Run the substitution backwards