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Difference between revisions of "2017 AIME II Problems/Problem 6"

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==Solution 1==
 
==Solution 1==
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Manipulating the given expression, <math>\sqrt{n^2+85n+2017}=\frac{1}{2}\sqrt{4n^2+340n+8068}=\frac{1}{2}\sqrt{(2n+85)^2+843}</math>. The expression under the radical must be an square number for the entire expression to be an integer, so <math>(2n+85)^2+843=s^2</math>. Rearranging, <math>s^2-(2n+85)^2=843</math>. By difference of squares, <math>(s-(2n+85))(s+(2n+85))=1\times843=3\times281</math>. It is easy to check that those are all the factor pairs of 843. Considering each factor pair separately, <math>2n+85</math> is found to be <math>421</math> and <math>139</math>. The two values of <math>n</math> that satisfy one of the equations are <math>168</math> and <math>27</math>. Summing these together, the answer is <math>168+27=\boxed{195}</math>.
+
Manipulating the given expression, <math>\sqrt{n^2+85n+2017}=\frac{1}{2}\sqrt{4n^2+340n+8068}=\frac{1}{2}\sqrt{(2n+85)^2+843}</math>. The expression under the radical must be an square number for the entire expression to be an integer, so <math>(2n+85)^2+843=s^2</math>. Rearranging, <math>s^2-(2n+85)^2=843</math>. By difference of squares, <math>(s-(2n+85))(s+(2n+85))=1\times843=3\times281</math>. It is easy to check that those are all the factor pairs of 843. Considering each factor pair separately, <math>2n+85</math> is found to be <math>421</math> and <math>139</math>. The two values of <math>n</math> that satisfy one of the equations are <math>168</math> and <math>27</math>. Summing these together gives us the answer ; <math>168+27=\boxed{195}</math>.
  
 
==Solution 2==
 
==Solution 2==
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Since <math>n</math> is positive, there are two cases, which are simple (luckily). Remembering that <math>x</math> is a positive integer, then <math>x^2-2017</math> and <math>85-2x</math> are either both positive or both negative. The smallest value for which <math>x^2>2017</math> is 45, which makes the denominator, and the entire expression, negative. Evaluating the other case where numerator and denominator are both negative, then we have that <math>x<45</math> (from the numerator) and <math>85-2x<0</math>, which means <math>x>42</math>. This only gives two solutions, <math>x=43, 44</math>. Plugging these into the expression for <math>n</math>, we find that they result in 27 and 168, which both satisfy the initial question. Therefore, the answer is <math>168+27=\boxed{195}</math>.
 
Since <math>n</math> is positive, there are two cases, which are simple (luckily). Remembering that <math>x</math> is a positive integer, then <math>x^2-2017</math> and <math>85-2x</math> are either both positive or both negative. The smallest value for which <math>x^2>2017</math> is 45, which makes the denominator, and the entire expression, negative. Evaluating the other case where numerator and denominator are both negative, then we have that <math>x<45</math> (from the numerator) and <math>85-2x<0</math>, which means <math>x>42</math>. This only gives two solutions, <math>x=43, 44</math>. Plugging these into the expression for <math>n</math>, we find that they result in 27 and 168, which both satisfy the initial question. Therefore, the answer is <math>168+27=\boxed{195}</math>.
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==Solution 3 (Abuse the discriminant)==
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Let the integer given by the square root be represented by <math>x</math>. Then <math>0 = n^2 + 85n + 2017 - x^2</math>. For this to have rational solutions for <math>n</math> (checking whether they are integers is done later), the discriminant of this quadratic must be a perfect square. (This can be easily shown using the quadratic formula.)
 +
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Thus, <math>b^2 - 4ac = 7225 + 4x^2 - 8068 = y^2</math> for some integer <math>y</math>. Then <math>4x^2 - 843 = y^2</math>. Rearranging this equation yields that <math>843 = (2x+y)(2x-y)</math>. Noticing that there are 2 factor pairs of <math>843</math>, namely, <math>1*843</math> and <math>3*281</math>, there are 2 systems to solve for <math>x</math> and <math>y</math> that create rational <math>n</math>. These yield solutions <math>(x,y)</math> of <math>(211, 421)</math> and <math>(71, 139)</math>.
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The solution to the initial quadratic in <math>n</math> must then be <math>\frac{-85 \pm \sqrt{85^2 - 4(2017 - x^2)}}{2}</math>. Noticing that for each value of <math>x</math> that has rational solutions for <math>n</math>, the corresponding value of the square root of the discriminant is <math>y</math>, the formula can be rewritten as <math>n = \frac{-85 \pm y}{2}</math>. One solution is <math>\frac{421 - 85}{2} = 168</math> and the other solution is <math>\frac{139 - 85}{2} = 27</math>. Thus the answer is <math>168 + 27 = \boxed{195}</math> as both rational solutions are integers.
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==Solution 4 (Squeezing/Sandwich method)==
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Notice that <math>(n+42)^2= n^2+84n+1764</math>. Also note that <math>(n+45)^2= n^2+90n+2025</math>. Thus, <cmath>(n+42)^2< n^2+85n+2017<(n+45)^2</cmath> where <math>n^2+85n+2017</math> is a perfect square. Hence,<cmath>n^2+85n+2017= (n+43)^2</cmath> or <cmath>n^2+85n+2017= (n+44)^2.</cmath> Solving the two equations yields the two solutions <math>n= 168, 27</math>. Therefore, our answer is <math>\boxed{195}</math>.
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==Solution 5 (Using factors)==
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Let the expression be equal to <math>a</math>. This expression can be factored into <math>\sqrt{(n+44)^2-3n+81}</math>. Then square both sides, and the expression becomes <math>(n+44)^2-3n+81=a^2</math>. We have a difference of two squares. Rearranging, we have <math>(n+44+a)(n+44-a)=3(n-27)</math>. By inspection, the only possible values for <math>(n+44-a)</math> are 0 and 1. When <math>(n+44-a)=0</math>, we must have <math>n-27=0</math>. Therefore, <math>27</math> is a solution. When we have <math>(n+44-a)=1</math>, so <math>n=a-43</math>. Plugging this back to <math>(n+44+a)=3(n-27)</math> (since <math>(n+44-a)=1</math>), we find that <math>a=211 \implies n=168</math>. Thus, the answer is <math>27+168= \boxed{195}</math>.
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'''-RootThreeOverTwo'''
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==Solution 6==
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 +
Ignore the square root for now. This expression can be factored into <math>(n+44)^2-3n+81</math>. Just by inspection, when <math>n=27</math>, the expression becomes <math>71^2</math>, so <math>27</math> is a solution. Proceed as Solution 5 to find the other solution(s).
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 +
==Solution 7 (alternative factoring)==
 +
More intuitive, but a little bit slower considering the decimals.
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 +
 +
Label the entire given expression as <math>k^2</math>.
 +
 +
Instinctively we can do a crude completion of the square, resulting in <math>k^2</math> = <math>(n+42.5)^2+210.75</math>
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Rearrange the equation to get a difference of squares.
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<math>k^2-(n+42.5)^2 = 210.75</math>
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<math>(k+n+42.5)(k-n-42.5) = 210.75</math>
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Factor <math>21,075</math> to get <math>3^1</math>,<math>5^2</math>, and <math>281^1</math>
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Now the two factors given are either divided by 10 each or one being divided by 100. Let's start with the former case.
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 +
If you try <math>281*3/10</math> and <math>5*5/10</math>, you quickly realize that <math>n</math> becomes negative. Naturally, you will realize you want the number's difference to be larger. Try <math>281*5/10</math> and <math>3*5/10</math>. This gives an answer of <math>27</math> for <math>x</math>. The next largest possibility also works, giving an <math>n</math> of <math>168</math>. As you rise, some numbers don't work because it results in an n that is not an integer, as in the example of <math>\frac{281*5*5}{10}</math> and <math>\frac{3}{10}</math>.
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 +
Now if you continue on with the next case, where one factor is divided by <math>100</math>, very swiftly will you realize most don't work simply because the difference is too small, or it doesn't give an integer. It helps a lot when you realize that the decimal does not end in a <math>5</math>, the answer will not be an integer. After a few short tests, we get <math>168+27=\boxed{195}</math>.
 +
 +
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 +
-jackshi2006
 +
- minor latex edits by jske25
 +
 +
==Video Solution==
 +
 +
https://youtu.be/Z23Yz05eblY
  
 
=See Also=
 
=See Also=
 
{{AIME box|year=2017|n=II|num-b=5|num-a=7}}
 
{{AIME box|year=2017|n=II|num-b=5|num-a=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 09:04, 3 July 2024

Problem

Find the sum of all positive integers $n$ such that $\sqrt{n^2+85n+2017}$ is an integer.

Solution 1

Manipulating the given expression, $\sqrt{n^2+85n+2017}=\frac{1}{2}\sqrt{4n^2+340n+8068}=\frac{1}{2}\sqrt{(2n+85)^2+843}$. The expression under the radical must be an square number for the entire expression to be an integer, so $(2n+85)^2+843=s^2$. Rearranging, $s^2-(2n+85)^2=843$. By difference of squares, $(s-(2n+85))(s+(2n+85))=1\times843=3\times281$. It is easy to check that those are all the factor pairs of 843. Considering each factor pair separately, $2n+85$ is found to be $421$ and $139$. The two values of $n$ that satisfy one of the equations are $168$ and $27$. Summing these together gives us the answer ; $168+27=\boxed{195}$.

Solution 2

Clearly, the result when $n$ is plugged into the given expression is larger than $n$ itself. Let $x$ be the positive difference between that result and $n$, so that $\sqrt{n^2+85n+2017}=n+x$. Squaring both sides and canceling the $n^2$ terms gives $85n+2017=2xn+x^2$. Combining like terms, $(85-2x)n=x^2-2017$, so

\[n=\frac{x^2-2017}{85-2x}.\]

Since $n$ is positive, there are two cases, which are simple (luckily). Remembering that $x$ is a positive integer, then $x^2-2017$ and $85-2x$ are either both positive or both negative. The smallest value for which $x^2>2017$ is 45, which makes the denominator, and the entire expression, negative. Evaluating the other case where numerator and denominator are both negative, then we have that $x<45$ (from the numerator) and $85-2x<0$, which means $x>42$. This only gives two solutions, $x=43, 44$. Plugging these into the expression for $n$, we find that they result in 27 and 168, which both satisfy the initial question. Therefore, the answer is $168+27=\boxed{195}$.

Solution 3 (Abuse the discriminant)

Let the integer given by the square root be represented by $x$. Then $0 = n^2 + 85n + 2017 - x^2$. For this to have rational solutions for $n$ (checking whether they are integers is done later), the discriminant of this quadratic must be a perfect square. (This can be easily shown using the quadratic formula.)

Thus, $b^2 - 4ac = 7225 + 4x^2 - 8068 = y^2$ for some integer $y$. Then $4x^2 - 843 = y^2$. Rearranging this equation yields that $843 = (2x+y)(2x-y)$. Noticing that there are 2 factor pairs of $843$, namely, $1*843$ and $3*281$, there are 2 systems to solve for $x$ and $y$ that create rational $n$. These yield solutions $(x,y)$ of $(211, 421)$ and $(71, 139)$.

The solution to the initial quadratic in $n$ must then be $\frac{-85 \pm \sqrt{85^2 - 4(2017 - x^2)}}{2}$. Noticing that for each value of $x$ that has rational solutions for $n$, the corresponding value of the square root of the discriminant is $y$, the formula can be rewritten as $n = \frac{-85 \pm y}{2}$. One solution is $\frac{421 - 85}{2} = 168$ and the other solution is $\frac{139 - 85}{2} = 27$. Thus the answer is $168 + 27 = \boxed{195}$ as both rational solutions are integers.

Solution 4 (Squeezing/Sandwich method)

Notice that $(n+42)^2= n^2+84n+1764$. Also note that $(n+45)^2= n^2+90n+2025$. Thus, \[(n+42)^2< n^2+85n+2017<(n+45)^2\] where $n^2+85n+2017$ is a perfect square. Hence,\[n^2+85n+2017= (n+43)^2\] or \[n^2+85n+2017= (n+44)^2.\] Solving the two equations yields the two solutions $n= 168, 27$. Therefore, our answer is $\boxed{195}$.

Solution 5 (Using factors)

Let the expression be equal to $a$. This expression can be factored into $\sqrt{(n+44)^2-3n+81}$. Then square both sides, and the expression becomes $(n+44)^2-3n+81=a^2$. We have a difference of two squares. Rearranging, we have $(n+44+a)(n+44-a)=3(n-27)$. By inspection, the only possible values for $(n+44-a)$ are 0 and 1. When $(n+44-a)=0$, we must have $n-27=0$. Therefore, $27$ is a solution. When we have $(n+44-a)=1$, so $n=a-43$. Plugging this back to $(n+44+a)=3(n-27)$ (since $(n+44-a)=1$), we find that $a=211 \implies n=168$. Thus, the answer is $27+168= \boxed{195}$.

-RootThreeOverTwo

Solution 6

Ignore the square root for now. This expression can be factored into $(n+44)^2-3n+81$. Just by inspection, when $n=27$, the expression becomes $71^2$, so $27$ is a solution. Proceed as Solution 5 to find the other solution(s).

Solution 7 (alternative factoring)

More intuitive, but a little bit slower considering the decimals.


Label the entire given expression as $k^2$.

Instinctively we can do a crude completion of the square, resulting in $k^2$ = $(n+42.5)^2+210.75$ Rearrange the equation to get a difference of squares.

$k^2-(n+42.5)^2 = 210.75$

$(k+n+42.5)(k-n-42.5) = 210.75$

Factor $21,075$ to get $3^1$,$5^2$, and $281^1$

Now the two factors given are either divided by 10 each or one being divided by 100. Let's start with the former case.

If you try $281*3/10$ and $5*5/10$, you quickly realize that $n$ becomes negative. Naturally, you will realize you want the number's difference to be larger. Try $281*5/10$ and $3*5/10$. This gives an answer of $27$ for $x$. The next largest possibility also works, giving an $n$ of $168$. As you rise, some numbers don't work because it results in an n that is not an integer, as in the example of $\frac{281*5*5}{10}$ and $\frac{3}{10}$.

Now if you continue on with the next case, where one factor is divided by $100$, very swiftly will you realize most don't work simply because the difference is too small, or it doesn't give an integer. It helps a lot when you realize that the decimal does not end in a $5$, the answer will not be an integer. After a few short tests, we get $168+27=\boxed{195}$.


-jackshi2006 - minor latex edits by jske25

Video Solution

https://youtu.be/Z23Yz05eblY

See Also

2017 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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