AMC 10 · 2020 · #24

Grade 8 arithmetic

Archetype Small-Domain Constrained Search

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

📏 Long solution 💡 3 insights

Problem

How many positive integers $n$ satisfy $\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor?$ (Recall that $\lfloor x\rfloor$ is the greatest integer not exceeding $x$ .)

$\textbf{(A) } 2 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 30 \qquad\textbf{(E) } 32$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

View mode:

Toolkit + CCSS Solution

Understand

Restated: Count the positive integers $n$ for which $\dfrac{n + 1000}{70} \;=\; \lfloor \sqrt{n} \rfloor$, where $\lfloor x \rfloor$ is the greatest integer $\le x$.

Givens: Equation $\dfrac{n + 1000}{70} = \lfloor \sqrt{n} \rfloor$ for positive integers $n$; Both sides must be the same integer; Answer choices: (A) $2$, (B) $4$, (C) $6$, (D) $30$, (E) $32$

Unknowns: The number of positive integers $n$ satisfying the equation

Understand

Restated: Count the positive integers $n$ for which $\dfrac{n + 1000}{70} \;=\; \lfloor \sqrt{n} \rfloor$, where $\lfloor x \rfloor$ is the greatest integer $\le x$.

Givens: Equation $\dfrac{n + 1000}{70} = \lfloor \sqrt{n} \rfloor$ for positive integers $n$; Both sides must be the same integer; Answer choices: (A) $2$, (B) $4$, (C) $6$, (D) $30$, (E) $32$

Plan

Primary tool: #13 Convert to Algebra

Secondary: #7 Identify Subproblems, #6 Guess and Check, #2 Make a Systematic List, #3 Eliminate Possibilities

Tool #13 (Algebra): introduce $k = \lfloor \sqrt{n} \rfloor$ and rewrite as $n = 70k - 1000$. Tool #7 (Subproblems): the floor condition $k^2 \le n < (k+1)^2$ 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 $n \le 0$.

Execute — Answer: C

#13 Convert to Algebra 6.EE.B.7 Step 1

Let $k = \lfloor \sqrt{n} \rfloor$.
Both sides of the equation are integers, so $\dfrac{n + 1000}{70} = k$ gives $n = 70k - 1000$.

$$k = \lfloor \sqrt{n} \rfloor, \;\; n = 70k - 1000$$

💡 Introduce a single name $k$ for both sides; the equation becomes linear in $n$ for each $k$.

#7 Identify Subproblems 8.EE.C.7 Step 2

The defining inequality for $k = \lfloor \sqrt{n} \rfloor$ is $k^2 \le n < (k+1)^2$, which $n$ must satisfy.
Plug in $n = 70k - 1000$: $$k^2 \;\le\; 70k - 1000 \;<\; (k+1)^2 \;=\; k^2 + 2k + 1.$$

$$k^2 \le 70k - 1000 < k^2 + 2k + 1$$

💡 Two simultaneous inequalities — one for the lower edge of the floor, one for the upper edge.

#13 Convert to Algebra 8.EE.C.7 Step 3

Left inequality: $k^2 - 70k + 1000 \le 0$.
Solve by quadratic formula: roots $k = \dfrac{70 \pm \sqrt{4900 - 4000}}{2} = \dfrac{70 \pm 30}{2} = 20 \text{ or } 50$.
So $20 \le k \le 50$.

$$k^2 - 70k + 1000 \le 0 \Leftrightarrow 20 \le k \le 50$$

💡 Parabola opens up; it's $\le 0$ between its two roots $20$ and $50$.

#13 Convert to Algebra 8.EE.C.7 Step 4

Right inequality: $70k - 1000 < k^2 + 2k + 1$, i.e., $k^2 - 68k + 1001 > 0$.
Roots: $k = \dfrac{68 \pm \sqrt{4624 - 4004}}{2} = \dfrac{68 \pm \sqrt{620}}{2}$.
Numerically $\sqrt{620} \approx 24.90$, so roots $\approx \dfrac{68 \pm 24.90}{2} \approx 21.55, 46.45$.
The parabola opens up, so $> 0$ outside $(21.55, 46.45)$: $k \le 21$ or $k \ge 47$.

$$k^2 - 68k + 1001 > 0 \Leftrightarrow k \le 21 \text{ or } k \ge 47$$

💡 Parabola is positive outside its roots — split into two pieces.

#2 Make a Systematic List 6.EE.B.8 Step 5

Intersect the two constraints: $\{20 \le k \le 50\} \cap \{k \le 21 \text{ or } k \ge 47\} = \{20, 21\} \cup \{47, 48, 49, 50\}$.
Six integer values total.

$$k \in \{20, 21, 47, 48, 49, 50\}$$

💡 Two short integer runs at the edges of $[20, 50]$.

#6 Guess and Check 8.NS.A.2 Step 6

Each $k$ in the list gives a positive $n = 70k - 1000 > 0$ (since $k \ge 20$, $n \ge 400 > 0$).
Now verify the original equation for each $k$ by checking $\lfloor \sqrt{n} \rfloor = k$: • $k = 20$: $n = 400$, $\sqrt{400} = 20$, floor $= 20$ ✓ • $k = 21$: $n = 470$, $\sqrt{470} \approx 21.68$, floor $= 21$ ✓ • $k = 47$: $n = 2290$, $\sqrt{2290} \approx 47.85$, floor $= 47$ ✓ • $k = 48$: $n = 2360$, $\sqrt{2360} \approx 48.58$, floor $= 48$ ✓ • $k = 49$: $n = 2430$, $\sqrt{2430} \approx 49.30$, floor $= 49$ ✓ • $k = 50$: $n = 2500$, $\sqrt{2500} = 50$, floor $= 50$ ✓

$$n \in \{400, 470, 2290, 2360, 2430, 2500\}$$

💡 Plug each $k$ back and read off $\sqrt{n}$ numerically to confirm the floor.

#3 Eliminate Possibilities 6.EE.B.5 Step 7

All six $n$ values check out.
So the count is $6$, matching choice $(C)$.

$$\#\{n\} = 6 \Rightarrow \textbf{(C)}$$

💡 Six solutions confirmed — answer $(C)$.

[1] #13 6.EE.B.7 Let $k = \lfloor \sqrt{n} \rfloor$. Both sides of the equation are integers, so

[2] #7 8.EE.C.7 The defining inequality for $k = \lfloor \sqrt{n} \rfloor$ is $k^2 \le n < (k+1)

[3] #13 8.EE.C.7 Left inequality: $k^2 - 70k + 1000 \le 0$. Solve by quadratic formula: roots $k

[4] #13 8.EE.C.7 Right inequality: $70k - 1000 < k^2 + 2k + 1$, i.e., $k^2 - 68k + 1001 > 0$. Roo

[5] #2 6.EE.B.8 Intersect the two constraints: ${20 \le k \le 50} \cap {k \le 21 \text{ or }

[6] #6 8.NS.A.2 Each $k$ in the list gives a positive $n = 70k - 1000 > 0$ (since $k \ge 20$, $n

[7] #3 6.EE.B.5 All six $n$ values check out. So the count is $6$, matching choice $(C)$.

Review

Reasonableness: 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 $0$). $k = 20$: $n = 400 = 20^2$ — exact match. $k = 50$: $n = 2500 = 50^2$ — 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 $22 \le k \le 46$ fails the right inequality because $(k+1)^2 - k^2 = 2k + 1$ is too narrow to contain the linear $70k - 1000 - k^2$ when $k$ is mid-range. Six solutions matches choice $(C)$.

Alternative: Tool #2 (Systematic List) directly: $n + 1000$ must be a multiple of $70$, so $n \in \{40, 110, 180, \dots\}$ (arithmetic progression with common difference $70$). For each such $n$, compute $\lfloor \sqrt{n} \rfloor$ and check equality with $(n + 1000)/70$. This loops over $\sim 50$ candidates and is concrete but slower than the algebraic approach.

CCSS standards used (min grade 8)

6.EE.B.5 Understand solving an equation or inequality as a process of finding values (Final count of integer solutions — collecting all $k$ that satisfy both inequalities and the floor check.)
6.EE.B.7 Solve real-world problems by writing and solving equations of the form px = q (Rearranging $(n + 1000)/70 = k$ to $n = 70k - 1000$.)
6.EE.B.8 Write an inequality of the form x > c or x < c and graph on a number line (Intersecting the integer intervals $\{20 \le k \le 50\}$ and $\{k \le 21 \text{ or } k \ge 47\}$.)
8.EE.C.7 Solve linear equations in one variable (Solving the quadratic inequalities $k^2 - 70k + 1000 \le 0$ and $k^2 - 68k + 1001 > 0$ via the quadratic formula.)
8.NS.A.2 Use rational approximations of irrational numbers to compare their size (Approximating $\sqrt{620} \approx 24.90$ and the various $\sqrt{n}$ values to verify each floor.)

⭐ This AMC 10 problem only needs Grade 8 inequalities — substitute $k = \lfloor \sqrt{n} \rfloor$ to get $n = 70k - 1000$, plug into the floor's defining inequalities $k^2 \le n < (k+1)^2$, solve two quadratics, and intersect: $k \in \{20, 21, 47, 48, 49, 50\}$ gives $6$ solutions.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 8)

More like this