2024 AMC 10B Problems/Problem 13: Difference between revisions

Latest revision as of 18:53, 10 November 2025

Problem

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

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

Solution 1

Note that $\sqrt{1183}=13\sqrt7$ . Since $x$ and $y$ are positive integers, and $\sqrt{x}+\sqrt{y}=\sqrt{1183}$ we can represent each value of $\sqrt{x}$ and $\sqrt{y}$ as the product of a positive integer and $\sqrt7$ . Let's say that $\sqrt{x}=m\sqrt7$ and $\sqrt{y}=n\sqrt7$ , where $m$ and $n$ are positive integers. This implies that $x+y=(\sqrt{x})^2+(\sqrt{y})^2=7m^2+7n^2=7(m^2+n^2)$ and that $m+n=13$ . WLOG, assume that ${m}\geq{n}$ . It is not hard to see that $x+y$ reaches its minimum when $m^2+n^2$ reaches its minimum. We now apply algebraic manipulation to get that $m^2+n^2=(m+n)^2-2mn$ Since $m+n$ is determined, we now want $mn$ to reach its maximum. Since $m$ and $n$ are positive integers, we can use the AM-GM inequality to get that: $\frac{m+n}{2}\geq{\sqrt{mn}}$ . When $mn$ reaches its maximum, $\frac{m+n}{2}={\sqrt{mn}}$ . This implies that $m=n=\frac{13}{2}$ . However, this is not possible since $m$ and $n$ and integers. Under this constraint, we can see that $mn$ reaches its maximum when $m=7$ and $n=6$ . Therefore, the minimum possible value of $x+y$ is $7(m^2+n^2)=7(7^2+6^2)=\boxed{\textbf{(B)}595}$

~Bloggish

A similar method is to take $y=1183-26\sqrt{7x}-x^2$ , then noting $x=7a^2$ and bashing to find the value of a where x is closest to y.

~meihk_neiht

Solution 2 (Guessing & Answer Choices)

Set $x=y$ , giving the minimum possible values. The given equation becomes $\sqrt{x}=\sqrt{y}=\dfrac{\sqrt{1183}}{2}\implies x=y=\dfrac{1183}{4}.$ This means that $x+y=\dfrac{1183}{2}=591.5.$ Since this is closest to answer choice $\text{(B)}$ , the answer is $\boxed{\textbf{(B) }595}$ ~Neoronean ~Tacos_are_yummy_1 (latex)

Note: If using a solution similar to this one it is recommended to still find valid x and y that add to 595.

~meikh neiht

Solution 3 (quick, no guessing but still answer choices)

Square both sides of the equation to get $2\cdot \sqrt{xy} + (x + y) = 1183.$ We can plug in possible values for $x + y$ based on the answer choices. If we do $585,$ we get $\sqrt{xy} = 299 \implies xy = 299^2.$ So $y = \frac{299^2}{x}.$ Then we have $x^2 + 299^2 = 585x,$ which has no real solutions for $x$ since the discriminant is negative. Following the same process for $595$ we obtain the equation $x^2 + 294^2 = 595x,$ which can be factored as $(x - 252)(x - 343) = 0.$ So the answer must be $\boxed{\textbf{(B) }595}$ .

~grogg007

Solution 4 (Calculus)

Since $\sqrt{x}+\sqrt{y}=\sqrt{1183}$ , we can express $y$ in terms of $x$ : $y = x+1183-2\sqrt{1183x}$ . $x+y = 2x+1183-2\sqrt{1183x}$ . Taking the derivative of this, we get $2-2 \cdot \frac{1}{2\sqrt{1183x}} \cdot 1183$ . We find the critical points by setting this expression to equate $0$ . Simplifying, we get $2 = \frac{1183}{\sqrt{1183x}}$ , and then we get $4 \cdot 1183x = 1183^2$ . Dividing by $1183$ , we get $4x=1183$ , so a critical point is $x=\frac{1183}{4}$ . Now, we take the second derivative, we get $\frac{13\sqrt{7x}}{2x^2}$ . When $x=\frac{1183}{4}$ , the second derivative is positive, meaning the minimum value is obtained at $x=\frac{1183}{4}$ . This value is $295.75$ . However, the problem asks for positive integers $x$ and $y$ . We want $x$ to be as close to $295.75$ as possible. We also want $y$ to be an integer, so $x+1183-2\sqrt{1183x}$ should be an integer. Notice how since $x$ is an integer, we just want $\sqrt{1183x}$ to be an integer. Prime factorizing $1183$ , we get $1183=7 \cdot 13^2$ . Therefore, $x$ must contain an odd amount of $7$ 's and an even amount of any other prime factor. We quickly find that $2^2 \cdot 3^2 \cdot 7 = 252$ is the optimal solution for $x$ , and plugging in, we get $y=252 + 1183 - 2 \cdot 2 \cdot 3 \cdot 7 \cdot 13 = 343$ . We find $x+y=595$ , so the answer is choice $\boxed{\text{(B) 595}}$ . ~ lovelearning999

Note: It might not be immediately clear why $x=252$ is the optimal solution. The following explains why $x=252$ is optimal.

$x$ must have an odd amount of $7$ 's and an even amount of any other prime factor. This means that $x$ can be expressed as $7^n \cdot a$ where $a$ is a perfect square. We can simplify this to $7 \cdot 7^{n-1} \cdot a = 7b$ , where $b$ is a perfect square, since $n-1$ is even. This means that $7b$ is as close to $295.75$ as possible. This means that $b \approx \frac{295.75}{7}$ . $\frac{295.75}{7}$ is approximately $42$ , so $b \approx 42$ . Since $b$ is a perfect square, we have two candidates for $b$ : $36$ and $49$ . If $b=36$ , $x=252$ , and if $b=49$ , $x=343$ . To find the optimal value for $x$ , pick the value closest to $295.75$ , and we find that $\left |295.75-252 \right | < \left |295.75 - 343 \right|$ , so $x=252$ is optimal. From now, proceed in the same way as the original solution.

~ lovelearning999

Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)

https://youtu.be/YqKmvSR1Ckk?feature=shared

~ Pi Academy

Video Solution 2 by SpreadTheMathLove

https://youtu.be/7ZKvxU6c75g?si=KvM1x612u6fig_Xc

Video Solution 3 by TheBeautyofMath

https://youtu.be/dfF39udgqc8?t=724 in Rapid Fire

~IceMatrix

@@ Line 5: / Line 5: @@
 ==Solution 1==
-Note that <math>\sqrt{1183}=13\sqrt7</math>. Since <math>x</math> and <math>y</math> are positive integers, and <math>\sqrt{x}+\sqrt{y}=\sqrt{1183}</math>, we can represent each value of <math>\sqrt{x}</math> and <math>\sqrt{y}</math> as the product of a positive integer and <math>\sqrt7</math>. Let's say that <math>\sqrt{x}=m\sqrt7</math> and <math>\sqrt{y}=n\sqrt7</math>, where <math>m</math> and <math>n</math> are positive integers. This implies that <math>x+y=(\sqrt{x})^2+(\sqrt{y})^2=7m^2+7n^2=7(m^2+n^2)</math> and that <math>m+n=13</math>. WLOG, assume that <math>{m}\geq{n}</math>. It is not hard to see that <math>x+y</math> reaches its minimum when <math>m^2+n^2</math> reaches its minimum. We now apply algebraic manipulation to get that <math>m^2+n^2=(m+n)^2-2mn</math>. Since <math>m+n</math> is determined, we now want <math>mn</math> to reach its maximum. Since <math>m</math> and <math>n</math> are positive integers, we can use the AM-GM inequality to get that: <math>\frac{m+n}{2}\geq{\sqrt{mn}}</math>. When <math>mn</math> reaches its maximum, <math>\frac{m+n}{2}={\sqrt{mn}}</math>. This implies that <math>m=n=\frac{13}{2}</math>. However, this is not possible since <math>m</math> and <math>n</math> and integers. Under this constraint, we can see that <math>mn</math> reaches its maximum when <math>m=7</math> and <math>n=6</math>. Therefore, the minimum possible value of <math>x+y</math> is <math>7(m^2+n^2)=7(7^2+6^2)=\boxed{\textbf{(B)}595}</math>
+Note that <math>\sqrt{1183}=13\sqrt7</math>. Since <math>x</math> and <math>y</math> are positive integers, and <math>\sqrt{x}+\sqrt{y}=\sqrt{1183}</math> we can represent each value of <math>\sqrt{x}</math> and <math>\sqrt{y}</math> as the product of a positive integer and <math>\sqrt7</math>. Let's say that <math>\sqrt{x}=m\sqrt7</math> and <math>\sqrt{y}=n\sqrt7</math>, where <math>m</math> and <math>n</math> are positive integers. This implies that <cmath>x+y=(\sqrt{x})^2+(\sqrt{y})^2=7m^2+7n^2=7(m^2+n^2)</cmath> and that <math>m+n=13</math>. WLOG, assume that <math>{m}\geq{n}</math>. It is not hard to see that <math>x+y</math> reaches its minimum when <math>m^2+n^2</math> reaches its minimum. We now apply algebraic manipulation to get that <cmath>m^2+n^2=(m+n)^2-2mn</cmath> Since <math>m+n</math> is determined, we now want <math>mn</math> to reach its maximum. Since <math>m</math> and <math>n</math> are positive integers, we can use the AM-GM inequality to get that: <math>\frac{m+n}{2}\geq{\sqrt{mn}}</math>. When <math>mn</math> reaches its maximum, <math>\frac{m+n}{2}={\sqrt{mn}}</math>. This implies that <math>m=n=\frac{13}{2}</math>. However, this is not possible since <math>m</math> and <math>n</math> and integers. Under this constraint, we can see that <math>mn</math> reaches its maximum when <math>m=7</math> and <math>n=6</math>. Therefore, the minimum possible value of <math>x+y</math> is <math>7(m^2+n^2)=7(7^2+6^2)=\boxed{\textbf{(B)}595}</math>
 ~[[User:Bloggish|Bloggish]]
+A similar method is to take <math>y=1183-26\sqrt{7x}-x^2</math>, then noting <math>x=7a^2</math> and bashing to find the value of a where x is closest to y.
+~meihk_neiht
 ==Solution 2 (Guessing & Answer Choices)==
 Set <math>x=y</math>, giving the minimum possible values. The given equation becomes <cmath>\sqrt{x}=\sqrt{y}=\dfrac{\sqrt{1183}}{2}\implies x=y=\dfrac{1183}{4}.</cmath>This means that <cmath>x+y=\dfrac{1183}{2}=591.5.</cmath>Since this is closest to answer choice <math>\text{(B)}</math>, the answer is <math>\boxed{\textbf{(B) }595}</math> ~Neoronean ~Tacos_are_yummy_1 (latex)
+Note:
+If using a solution similar to this one it is recommended to still find valid x and y that add to 595.
+~meikh neiht
+==Solution 3 (quick, no guessing but still answer choices)==
+Square both sides of the equation to get <imath>2\cdot \sqrt{xy} + (x + y) = 1183.</imath> We can plug in possible values for <imath>x + y</imath> based on the answer choices. If we do <imath>585,</imath> we get <imath>\sqrt{xy} = 299 \implies xy = 299^2.</imath> So <imath>y = \frac{299^2}{x}.</imath> Then we have
+<imath>x^2 + 299^2 = 585x,</imath> which has no real solutions for <imath>x</imath> since the discriminant is negative. Following the same process for <imath>595</imath> we obtain the equation <imath>x^2 + 294^2 = 595x,</imath> which can be factored as <imath>(x - 252)(x - 343) = 0.</imath> So the answer must be <imath>\boxed{\textbf{(B) }595}</imath>.
+~[[User:grogg007|grogg007]]
+== Solution 4 (Calculus) ==
+Since <imath>\sqrt{x}+\sqrt{y}=\sqrt{1183}</imath>, we can express <imath>y</imath> in terms of <imath>x</imath>: <imath>y = x+1183-2\sqrt{1183x}</imath>. <imath>x+y = 2x+1183-2\sqrt{1183x}</imath>. Taking the derivative of this, we get <imath>2-2 \cdot \frac{1}{2\sqrt{1183x}} \cdot 1183</imath>. We find the critical points by setting this expression to equate <imath>0</imath>. Simplifying, we get <imath>2 = \frac{1183}{\sqrt{1183x}}</imath>, and then we get <imath>4 \cdot 1183x = 1183^2</imath>. Dividing by <imath>1183</imath>, we get <imath>4x=1183</imath>, so a critical point is <imath>x=\frac{1183}{4}</imath>. Now, we take the second derivative, we get <imath>\frac{13\sqrt{7x}}{2x^2}</imath>. When <imath>x=\frac{1183}{4}</imath>, the second derivative is positive, meaning the minimum value is obtained at <imath>x=\frac{1183}{4}</imath>. This value is <imath>295.75</imath>. However, the problem asks for positive integers <imath>x</imath> and <imath>y</imath>. We want <imath>x</imath> to be as close to <imath>295.75</imath> as possible. We also want <imath>y</imath> to be an integer, so <imath>x+1183-2\sqrt{1183x}</imath> should be an integer. Notice how since <imath>x</imath> is an integer, we just want <imath>\sqrt{1183x}</imath> to be an integer. Prime factorizing <imath>1183</imath>, we get <imath>1183=7 \cdot 13^2</imath>. Therefore, <imath>x</imath> must contain an odd amount of <imath>7</imath>'s and an even amount of any other prime factor. We quickly find that <imath>2^2 \cdot 3^2 \cdot 7 = 252</imath> is the optimal solution for <imath>x</imath>, and plugging in, we get <imath>y=252 + 1183 - 2 \cdot 2 \cdot 3 \cdot 7 \cdot 13 = 343</imath>. We find <imath>x+y=595</imath>, so the answer is choice <imath>\boxed{\text{(B) 595}}</imath>.
+~ lovelearning999
+Note:
+It might not be immediately clear why <imath>x=252</imath> is the optimal solution. The following explains why <imath>x=252</imath> is optimal.
+<imath>x</imath> must have an odd amount of <imath>7</imath>'s and an even amount of any other prime factor. This means that <imath>x</imath> can be expressed as <imath>7^n \cdot a</imath> where <imath>a</imath> is a perfect square. We can simplify this to <imath>7 \cdot 7^{n-1} \cdot a = 7b</imath>, where <imath>b</imath> is a perfect square, since <imath>n-1</imath> is even. This means that <imath>7b</imath> is as close to <imath>295.75</imath> as possible. This means that <imath>b \approx \frac{295.75}{7}</imath>. <imath>\frac{295.75}{7}</imath> is approximately <imath>42</imath>, so <imath>b \approx 42</imath>. Since <imath>b</imath> is a perfect square, we have two candidates for <imath>b</imath>: <imath>36</imath> and <imath>49</imath>. If <imath>b=36</imath>, <imath>x=252</imath>, and if <imath>b=49</imath>, <imath>x=343</imath>. To find the optimal value for <imath>x</imath>, pick the value closest to <imath>295.75</imath>, and we find that <imath>\left |295.75-252 \right | < \left |295.75 - 343 \right|</imath>, so <imath>x=252</imath> is optimal. From now, proceed in the same way as the original solution.
+~ lovelearning999
 ==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)==
@@ Line 20: / Line 48: @@
 ==Video Solution 2 by SpreadTheMathLove==
 https://youtu.be/7ZKvxU6c75g?si=KvM1x612u6fig_Xc
+==Video Solution 3 by TheBeautyofMath==
+https://youtu.be/dfF39udgqc8?t=724 in Rapid Fire
+~IceMatrix
 ==See also==
 {{AMC10 box|year=2024|ab=B|num-b=12|num-a=14}}
 {{MAA Notice}}
+[[Category: Introductory Algebra Problems]]

2024 AMC 10B (Problems • Answer Key • Resources)
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