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Source: 2024 AMC 10B #13

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286 posts

#1 h Nov 13, 2024, 5:29 PM

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

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

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alexanderhamilton124

381 posts

alexanderhamilton124

#2 h Nov 13, 2024, 5:31 PM

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Answer is B) i think

use AM-Gm to eliminate A)

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164 posts

#3 h Nov 13, 2024, 5:31 PM

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B confirmed

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5310 posts

#4 h Nov 13, 2024, 5:32 PM

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B, outline sqrt(1183) = 13sqrt(7). wlog sqrt(x)= 7sqrt7 and sqrt(y) = 6sqrt7, solve

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pingpongmerrily

3518 posts

pingpongmerrily

#5 h Nov 13, 2024, 5:44 PM

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i didn't even solve it just guessed with am-gm

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635 posts

#6 h Nov 13, 2024, 5:51 PM

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yeah, just 6sqrt7 and 7sqrt7

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131 posts

#7 h Nov 13, 2024, 5:54 PM

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pretty sure i multiplied 85*7 wrong

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238 posts

Tem8

#8 h Nov 13, 2024, 5:55 PM

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$x+y=(\sqrt{x}+\sqrt{y})^2 - 2\sqrt{xy}=1183-2\sqrt{xy}.$ By $AM \ge GM$ , $\frac{\sqrt{x}+\sqrt{y}}{2} \ge \sqrt{\sqrt{xy}}.$ Squaring both sides, we find that
$\frac{1183}{4} \ge \sqrt{xy},$ so $-2\sqrt{xy} \ge -\frac{1183}{2}$ .
Thus, $x+y\ge1183-\frac{1183}{2}=591.5,$ so option A is eliminated automatically.
Now, let's check whether $x+y=595$ is possible for integer $x,y$ .

$\begin{align*} \sqrt{x}+\sqrt{y} &= \sqrt{1183}\\ x+y &= 595 \\ 2\sqrt{xy} &= (\sqrt{x}+\sqrt{y})^2-(x+y)=588\\ xy &= 294^2 \\ (x-y)^2 &= (x+y)^2 - 4xy \\ &= 595^2 - 2^2 \cdot 294^2 \\ &= 595^2 - 588^2 \\ &= (595-588)(595+588) \\ &=7\cdot1183=91^2 \\ x-y&=91\\ x&=\frac{595+91}{2},y=\frac{595-91}{2}. \end{align*}$

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5542 posts

#9 h Nov 13, 2024, 6:05 PM

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The answer is $\boxed{\textbf{(B)}\ 595}$ , achieved by $x = 343$ and $y = 252$ . Now we prove the bound.

Let $c = x + y$ . Squaring the equation gives $2\sqrt{xy} = 1183 - c$ , so $xy = \frac{(1183 - c)^2}{4}$ . Now considering the quadratic with roots $x,y$ , its discriminant, which is $(x+y)^2 - 4xy = c^2 - (1183 - c)^2 =1183 \cdot (2c - 1183) = 7 \cdot (2c - 1183) \cdot 13^2$ must be a perfect square.

This implies that $7 \mid 2c - 1183$ , so $2c - 1183 \ge 7$ (as it must be odd), and therefore $c \ge 595$ , as desired.

(from here, a way to find working $x,y$ is by doing $\frac{c \pm \sqrt{7 \cdot (2c-1183) \cdot 13^2}}{2} = \frac{c \pm 91}{2}$ , giving $\{x,y\} = \{252,343\}$ )

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#10 h Nov 13, 2024, 6:08 PM

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Tem8 wrote:

$x+y=(\sqrt{x}+\sqrt{y})^2 - 2\sqrt{xy}=1183-2\sqrt{xy}.$ By $AM \ge GM$ , $\frac{\sqrt{x}+\sqrt{y}}{2} \ge \sqrt{\sqrt{xy}}.$ Squaring both sides, we find that
$\frac{1183}{4} \ge \sqrt{xy},$ so $-2\sqrt{xy} \ge -\frac{1183}{2}$ .
Thus, $x+y\ge1183-\frac{1183}{2}=591.5,$ so option A is eliminated automatically.
Now, let's check whether $x+y=595$ is possible for integer $x,y$ .

$\begin{align*} \sqrt{x}+\sqrt{y} &= \sqrt{1183}\\ x+y &= 595 \\ 2\sqrt{xy} &= (\sqrt{x}+\sqrt{y})^2-(x+y)=588\\ xy &= 294^2 \\ (x-y)^2 &= (x+y)^2 - 4xy \\ &= 595^2 - 2^2 \cdot 294^2 \\ &= 595^2 - 588^2 \\ &= (595-588)(595+588) \\ &=7\cdot1183=91^2 \\ x-y&=91\\ x&=\frac{595+91}{2},y=\frac{595-91}{2}. \end{align*}$

this was my sol !!

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293 posts

#11 h Nov 13, 2024, 6:12 PM

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zhoujef000 wrote:

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

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

Factoring out $1183$ yield s $7\cdot13\cdot13$ ; $\sqrt{x}$ $\sqrt{y}$ must be in the form $a\sqrt{b} + c\sqrt{b}$ , so we choose 7 to be B, then we know $(a+c)^2$ has to equal 169, and $a+c$ has to equal 13. Since x and y are both positive, integers, to minimize it we let $a = 6$ and $c = 7$ (Closest integers summing to 13), which will equal $7\sqrt{7}+6\sqrt{7} = 13\sqrt{7} = \sqrt{1183}$ so $x = 36 \cdot 7$ and $y = 49 \cdot 7$ , summing them results in $\textbf{(B)} 595$

This post has been edited 6 times. Last edited by lucaswujc, Nov 13, 2024, 6:16 PM

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525 posts

#12 h Nov 13, 2024, 6:33 PM

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B confirmed 1183=13^2*7
smallest x and y will clearly be closest to each other by concavity of the sqrt function
13+-1/2=6 and 7
(6^2+7^2)*7=595

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617 posts

eg4334

#13 h Nov 13, 2024, 7:08 PM

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$\sqrt{1183} = 13\sqrt{7}$ , and from here it is clear that $6\sqrt{7}$ and $7\sqrt{7}$ is optimal.

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4168 posts

#14 h Nov 13, 2024, 7:25 PM

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zhoujef000 wrote:

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

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

i took the 12b and didnt get this but here's a silly calculus confirmation

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2830 posts

#15 h Nov 13, 2024, 7:29 PM

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Solution

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628 posts

#16 h Nov 13, 2024, 7:32 PM • 1 Y

Y by Alex-131

When you are so depressed from problem 12 that you forgot, how to do these problems

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490 posts

#17 h Nov 13, 2024, 8:31 PM

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First, notice that $1183$ is divisible by $7.$ (this can be found from checking small primes to divide into 1183). So, $1183 = 7 \cdot 169 = 7 \cdot 13^{2}.$ So, if $x = 7a^{2},$ and $y = 7b^{2},$ then $a+b = 13.$ By AM-GM, the minimum is when $a=6, b=7,$ so we get $7 \cdot (6^{2}+7^{2}) = 595.$

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1725 posts

#18 h Nov 13, 2024, 9:40 PM

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NOOOOOOOOOOOO I GOT A

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#19 h Nov 13, 2024, 9:42 PM • 4 Y

Y by goldenuni678, Kempu33334, ranu540, Tem8

Why is everyone using am-gm. I factored it and got 13sqrt(7) and i just assumed that it would be 6sqrt(7) and 7sqrt(7).

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36 posts

#20 h Nov 13, 2024, 11:03 PM

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zhoujef000 wrote:

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

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

$1183 = 7\cdot13^2$

$\sqrt{1183} = 13\sqrt{7}$

Intuitively, $\sqrt{x} = 6\sqrt{7}, \sqrt{y} = 7\sqrt{7}$

Therefore, $x+y = 252 + 343 = \boxed{595}$

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MathPerson12321

3627 posts

MathPerson12321

#21 h Nov 13, 2024, 11:27 PM

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Alex-131 wrote:

B, outline sqrt(1183) = 13sqrt(7). wlog sqrt(x)= 7sqrt7 and sqrt(y) = 6sqrt7, solve

Yeah thats the best solution without AMGM cheese

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2264 posts

#22 h Nov 14, 2024, 5:28 AM

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carcool0 wrote:

Why is everyone using am-gm. I factored it and got 13sqrt(7) and i just assumed that it would be 6sqrt(7) and 7sqrt(7).

use am-gm to show it cant be lower

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