AMC 10 · 2020 · #24

Grade 8 arithmetic

Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination

floor-functionbound-inequality-then-enumeratedivisibility-rulesperfect-squares convert-to-algebrabound-inequality-then-enumeratecasework ↑ Prerequisites: floor-functiondivisibility-rules

📏 Long solution 💡 3 insights

Problem

Count the positive integers n for which

\frac{n + 1000}{70}

= ⌊ √(n) ⌋, where ⌊ x ⌋ is the greatest integer ≤ x.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Convert to Algebra

Tool #13 (Algebra): introduce k = ⌊ √(n) ⌋ and rewrite as n = 70k - 1000. Tool #7 (Subproblems): the floor condition k² ≤ n ＜ (k+1)² gives two quadratic inequalities in k — solve each separately. Tool #6 (Guess and Check): verify each candidate k by plugging back into the original equation. Tool #2 (Systematic List): enumerate the small set of integer k values satisfying both inequalities. Tool #3 (Eliminate): reject any k outside the valid range or giving n ≤ 0.

1STEP 1

Let k = ⌊ √(n) ⌋. Both sides are integers, so $\frac{n + 1000}{70}$ = k gives n = 70k - 1000.

k = ⌊ √(n) ⌋, n = 70k - 1000

2STEP 2

Plug n = 70k - 1000 into the floor's k² ≤ n ＜ (k+1)²: k² ≤ 70k - 1000 ＜ k² + 2k + 1.

k² ≤ 70k - 1000 ＜ k² + 2k + 1

3STEP 3

Left side k² - 70k + 1000 ≤ 0 has roots $\frac{70 ± 30}{2}$ = 20 and 50, so 20 ≤ k ≤ 50.

k² - 70k + 1000 ≤ 0 ⇔ 20 ≤ k ≤ 50

4STEP 4

Right side k² - 68k + 1001 ＞ 0 has roots ≈ 21.55 and 46.45, so k ≤ 21 or k ≥ 47.

k² - 68k + 1001 ＞ 0 ⇔ k ≤ 21 or k ≥ 47

5STEP 5

Intersecting 20 ≤ k ≤ 50 with k ≤ 21 or k ≥ 47 leaves k ∈ {20, 21, 47, 48, 49, 50}.

k ∈ {20, 21, 47, 48, 49, 50}

6STEP 6

Each k gives n = 70k - 1000 ＞ 0 and ⌊ √(n) ⌋ = k holds: n ∈ {400, 470, 2290, 2360, 2430, 2500}.

n ∈ {400, 470, 2290, 2360, 2430, 2500}

7STEP 7

All six n satisfy the equation, so the count is 6 — choice (C).

#{n} = 6 → (C)

Answer

Sanity. The equation has solutions only near the two roots k = 20 and k = 50 of the lower-edge equation (where the parabola is exactly 0). k = 20: n = 400 = 20² — exact match. k = 50: n = 2500 = 50² — also exact. Just past each end, the floor can still cover one or a few more k's before the right inequality bites. The middle range 22 ≤ k ≤ 46 fails the right inequality because (k+1)² - k² = 2k + 1 is too narrow to contain the linear 70k - 1000 - k² when k is mid-range. Six solutions matches choice (C).

💡Key takeaway

This AMC 10 problem only needs Grade 8 inequalities — substitute k = ⌊ √(n) ⌋ to get n = 70k - 1000, plug into the floor's defining inequalities k² ≤ n ＜ (k+1)², solve two quadratics, and intersect: k ∈ {20, 21, 47, 48, 49, 50} gives 6 solutions.

More like this