easy olympiad problem

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Middle School Math Grades 5-8, Ages 10-13, MATHCOUNTS, AMC 8

Grades 5-8, Ages 10-13, MATHCOUNTS, AMC 8

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Middle School Math Grades 5-8, Ages 10-13, MATHCOUNTS, AMC 8

Grades 5-8, Ages 10-13, MATHCOUNTS, AMC 8

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#1 h Apr 17, 2025, 2:00 PM

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Find all positive integer values of $x$ such that
$\sqrt{x - 2011} + \sqrt{2011 - x} + 10$ is an integer.

This post has been edited 2 times. Last edited by kjhgyuio, Apr 17, 2025, 2:01 PM
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#2 h Apr 17, 2025, 2:14 PM

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Solution

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#3 h Apr 17, 2025, 7:44 PM

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$x-2011$ and $2011-x$ are each other's negatives, and you can't take the square root of a negative number while having an integer solution. Therefore the only solution that would work is if they were both non-negative, and based on this they must be zero because one positive number will lead to one negative number. Since 0 is neither negative or positive it's what's under both square roots. Therefore, $\sqrt{x-2011} = 0$ . Squaring both sides gives $x-2011 = 0$ , hence $x = 2011$ . This is the only solution.

Edit: I know it's verbose but it's an "olympiad problem"

This post has been edited 2 times. Last edited by Charizard_637, Apr 21, 2025, 5:44 PM
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#4 h Apr 18, 2025, 1:25 AM

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Charizard_637 wrote:

$x-2011$ and $2011-x$ are each other's negatives, and you can't take the square root of a negative number while having an integer solution. Therefore the only solution that would work is if they were both non-negative, and based on this they must be zero because one positive number will lead to one negative number. Since 0 is neither negative or positive it's what's under both square roots. Therefore, $\sqrt{x-2011} = 0$ . Squaring both sides gives $x-2011 = 0$ , hence $x = 2011$ . This is the only solution.

Edit: I know it's verbose but it's an olympiad problem

you dont have to do this, this is enough

Notice that for the expression to be real, $x-2011\geq0$ and $2011-x\geq 0$ , otherwise the imaginary part of the expression would be positive. Thus no solutions other than $x=2011$ exist, and inspection gives that $x=2011$ works. Thus the answer is $x=2011$ .

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deduck

#5 h Apr 18, 2025, 1:59 AM

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this is a past amc8 problem

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#6 h Apr 18, 2025, 6:01 PM

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The both the roots there are reals only if x=2011

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#7 h Apr 22, 2025, 6:18 PM

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#8 h Apr 23, 2025, 2:32 PM

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deduck wrote:

this is a past amc8 problem

oh
my

:rotfl:

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