AMC 10 · 2020 · #21

Grade 8 number-theory

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 for which the number plus one thousand, divided by seventy, equals the floor of the number's square root.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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

Name the floor value

Express the number via the floor value.

k = ⌊ √(n) ⌋, n = 70k - 1000

2STEP 2

Write the definition as inequalities

The definition becomes two inequalities.

k² ≤ 70k - 1000 ＜ k² + 2k + 1

3STEP 3

Solve the first

The first gives a wide range.

k² - 70k + 1000 ≤ 0 ⇔ 20 ≤ k ≤ 50

4STEP 4

Solve the second

The second cuts out the middle.

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

5STEP 5

List the surviving values

List what survives.

k ∈ {20, 21, 47, 48, 49, 50}

6STEP 6

Convert back

Convert each back to the number.

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

7STEP 7

Count them

There are 6 of them.

#{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 12 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.