AMC 10 · 2024 · #13

Grade 8 algebra

Archetype Small-Domain Constrained Search

perfect-squaresprime-factorizationsystematic-enumeration convert-to-algebraidentify-subproblemssystematic-enumeration ↑ Prerequisites: perfect-squaresprime-factorizationlinear-equations-two-var

📏 Medium solution 💡 3 insights

Problem

Positive integers $x$ and $y$ satisfy the equation $\sqrt{x} + \sqrt{y} = \sqrt{1183}$ . What is the minimum possible value of $x+y$ ?

$\textbf{(A) } 585 \qquad\textbf{(B) } 595 \qquad\textbf{(C) } 623 \qquad\textbf{(D) } 700 \qquad\textbf{(E) } 791$

Pick an answer.

(A)

585

(B)

595

(C)

623

(D)

700

(E)

791

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Toolkit + CCSS Solution

Understand

Restated: Positive integers $x$ and $y$ satisfy $\sqrt{x} + \sqrt{y} = \sqrt{1183}$. Find the minimum value of $x + y$.

Givens: $x, y$ are positive integers; $\sqrt{x} + \sqrt{y} = \sqrt{1183}$; Answer choices: (A) $585$, (B) $595$, (C) $623$, (D) $700$, (E) $791$

Unknowns: The minimum possible value of $x + y$

Understand

Restated: Positive integers $x$ and $y$ satisfy $\sqrt{x} + \sqrt{y} = \sqrt{1183}$. Find the minimum value of $x + y$.

Givens: $x, y$ are positive integers; $\sqrt{x} + \sqrt{y} = \sqrt{1183}$; Answer choices: (A) $585$, (B) $595$, (C) $623$, (D) $700$, (E) $791$

Plan

Primary tool: #13 Convert to Algebra

Secondary: #7 Identify Subproblems, #2 Make a Systematic List

The equation has irrational square roots on both sides, so the standard move is Tool #13 (Convert to Algebra): factor $\sqrt{1183}$ to expose its irrational part. Once $\sqrt{1183} = 13\sqrt{7}$, the only way two positive-integer square roots can sum to $13\sqrt{7}$ is if each one is also of the form $\text{(integer)}\sqrt{7}$ — that's the subproblem split (Tool #7): first force the form $x = 7a^2$, $y = 7b^2$, then minimize. Tool #2 (Systematic List) handles the finite minimization: list every positive-integer pair $(a, b)$ with $a + b = 13$ and pick the one with the smallest $a^2 + b^2$. Classic "sum fixed, minimize sum-of-squares" — closest to equal wins.

Execute — Answer: B

#13 Convert to Algebra 6.NS.B.4 Step 1

Simplify $\sqrt{1183}$ by prime factoring $1183$.
Test small primes: $1183 / 7 = 169 = 13^2$, so $1183 = 7 \cdot 13^2$.
Therefore $\sqrt{1183} = 13\sqrt{7}$, and the equation becomes $\sqrt{x} + \sqrt{y} = 13\sqrt{7}$.

$$1183 = 7 \cdot 13^2 \;\Longrightarrow\; \sqrt{x} + \sqrt{y} = 13\sqrt{7}$$

💡 Pull every perfect-square factor out of the radical first — the irrational "flavor" of $\sqrt{7}$ is what really matters.

#13 Convert to Algebra 8.NS.A.1 Step 2

Determine the form of $x$ and $y$.
Since $\sqrt{x} + \sqrt{y} = 13\sqrt{7}$ with $x, y$ integers, both $\sqrt{x}$ and $\sqrt{y}$ must be integer multiples of $\sqrt{7}$ (any rational part would have to cancel, which is impossible since $\sqrt{7}$ is irrational).
So $x = 7a^2$ and $y = 7b^2$ for positive integers $a, b$, giving $\sqrt{x} = a\sqrt{7}$ and $\sqrt{y} = b\sqrt{7}$.

$$x = 7a^2, \;\; y = 7b^2 \;\Longrightarrow\; a\sqrt{7} + b\sqrt{7} = 13\sqrt{7} \;\Longrightarrow\; a + b = 13$$

💡 Two integer-square-roots summing to an irrational multiple of $\sqrt{7}$ forces both summands into the same "$\sqrt{7}$-shape" — Grade 8 irrationals refuse to mix.

#7 Identify Subproblems 6.EE.A.3 Step 3

Rewrite the objective.
Plug $x = 7a^2$, $y = 7b^2$ into $x + y$ and pull out $7$.

$$x + y = 7a^2 + 7b^2 = 7(a^2 + b^2)$$

💡 Factor out the common $7$ so the only thing to minimize is $a^2 + b^2$.

#2 Make a Systematic List 4.OA.B.4 Step 4

Minimize $a^2 + b^2$ subject to $a + b = 13$ with positive integers $a, b$.
The systematic list of pairs: $(1,12), (2,11), (3,10), (4,9), (5,8), (6,7)$.
Their sums of squares: $1+144=145$, $4+121=125$, $9+100=109$, $16+81=97$, $25+64=89$, $36+49=85$.
The minimum is $85$, attained at $(a, b) = (6, 7)$ (closest to equal).

$$\min_{a+b=13} (a^2 + b^2) = 6^2 + 7^2 = 36 + 49 = 85$$

💡 Walk the pairs in order — the smallest sum-of-squares with a fixed sum always sits where the two numbers are as close as possible.

#7 Identify Subproblems 5.NBT.B.5 Step 5

Plug back in.
The minimum $x + y = 7 \cdot 85 = 595$.
Sanity check: $x = 7 \cdot 36 = 252$, $y = 7 \cdot 49 = 343$, and $\sqrt{252} + \sqrt{343} = 6\sqrt{7} + 7\sqrt{7} = 13\sqrt{7} = \sqrt{1183}$.

$$x + y = 7 \cdot 85 = 595 \;\Rightarrow\; \textbf{(B)}$$

💡 Multiply through and double-check by plugging both values into the original equation — Grade 5 mental-arithmetic level.

[1] #13 6.NS.B.4 Simplify $\sqrt{1183}$ by prime factoring $1183$. Test small primes: $1183 / 7 =

[2] #13 8.NS.A.1 Determine the form of $x$ and $y$. Since $\sqrt{x} + \sqrt{y} = 13\sqrt{7}$ with

[3] #7 6.EE.A.3 Rewrite the objective. Plug $x = 7a^2$, $y = 7b^2$ into $x + y$ and pull out $7$

[4] #2 4.OA.B.4 Minimize $a^2 + b^2$ subject to $a + b = 13$ with positive integers $a, b$. The

[5] #7 5.NBT.B.5 Plug back in. The minimum $x + y = 7 \cdot 85 = 595$. Sanity check: $x = 7 \cdot

Review

Reasonableness: The answer $595$ sits squarely in the middle of the choices, which fits a minimum that is non-trivial but not at the extreme. Cross-check: the next pair $(a, b) = (5, 8)$ gives $7 \cdot (25 + 64) = 7 \cdot 89 = 623$, exactly choice (C); the pair $(4, 9)$ gives $7 \cdot 97 = 679$ (not on the menu), and $(3, 10)$ gives $7 \cdot 109 = 763$. The menu (A)$=585$, (B)$=595$, (C)$=623$, (E)$=791$ matches the actual sum-of-squares ladder very cleanly, with (B) being our minimum and (C) the runner-up — a strong sign the algebra is right.

Alternative: Tool #5 (Look for a Pattern) plus the AM-QM inequality: for any positive reals with $a+b$ fixed, $a^2 + b^2$ is minimized when $a = b$. Since $a$ and $b$ must be integers with $a+b = 13$ (odd), the best integer choice splits $13$ as evenly as possible, i.e. $\{6, 7\}$. This avoids the explicit list and lands on $(6,7)$ in one line. Same answer (B) $= 595$.

CCSS standards used (min grade 8)

6.NS.B.4 Find greatest common factor and least common multiple of two numbers (Prime-factoring $1183 = 7 \cdot 13^2$ to extract the perfect-square factor $13^2$ from under the radical.)
8.NS.A.1 Know that numbers that are not rational are called irrational numbers (Recognizing that $\sqrt{7}$ is irrational, so $\sqrt{x}$ and $\sqrt{y}$ each must be an integer multiple of $\sqrt{7}$ for their sum to equal $13\sqrt{7}$ — forcing $x = 7a^2$, $y = 7b^2$.)
6.EE.A.3 Apply the properties of operations to generate equivalent expressions (Rewriting $x + y = 7a^2 + 7b^2 = 7(a^2 + b^2)$ so the minimization reduces to $a^2 + b^2$.)
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite (Listing positive-integer pairs $(a, b)$ with $a + b = 13$ in order to pick the pair that minimizes $a^2 + b^2$.)
5.NBT.B.5 Fluently multiply multi-digit whole numbers (Computing $7 \cdot 85 = 595$ and verifying $252 + 343 = 595$ at the end.)

⭐ This AMC 10 problem only needs Grade 8 "irrationals don't mix" plus Grade 4 systematic pair-listing that you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 8)

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