Difference between revisions of "1971 Canadian MO Problems/Problem 6"

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=== Solution===
 
=== Solution===
  
Notice <math>n^{2} + 2n + 12 = (n+1)^{2} + 11</math>. For this expression to be equal to a multiple of 121,  <math>(n+1)^{2} + 11</math> would have to equal a number in the form <math>121x</math>. Now we have the equation <math>(n+1)^{2} + 11 = 121x</math>. Subtracting <math>11</math> from both sides and then factoring out <math>11</math> on the right hand side results in <math>(n+1)^{2} = 11(11x - 11)</math>. Now we can say <math>(n-1) = 11</math> and <math>(n-1) = 11x - 1</math>. Solving the first equation results in <math>n=10</math>. Plugging in <math>n=10</math> in the second equation and solving for <math>x</math>, <math>x = 12/11</math>. Since <math>12/11</math> *<math>121</math> is clearly not a multiple of 121,  <math>n^{2} + 2n + 12</math> can never be a multiple of 121.
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Notice <math>n^{2} + 2n + 12 = (n+1)^{2} + 11</math>. For this expression to be equal to a multiple of 121,  <math>(n+1)^{2} + 11</math> would have to equal a number in the form <math>121x</math>. Now we have the equation <math>(n+1)^{2} + 11 = 121x</math>. Subtracting <math>11</math> from both sides and then factoring out <math>11</math> on the right hand side results in <math>(n+1)^{2} = 11(11x - 1)</math>. Now we can say <math>(n+1) = 11</math> and <math>(n+1) = 11x - 1</math>. Solving the first equation results in <math>n=10</math>. Plugging in <math>n=10</math> in the second equation and solving for <math>x</math>, <math>x = 12/11</math>. Since <math>12/11</math> *<math>121</math> is clearly not a multiple of 121,  <math>n^{2} + 2n + 12</math> can never be a multiple of 121.
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=== Solution 2 ===
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Assume that <math>n^2+2n+12=121k</math> for some integer <math>k</math> then
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<cmath>n^2+2n+(12-121k)=0</cmath>
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<cmath>\begin{align*}
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x&=\frac{-2\pm\sqrt{4-4(12-121k)}}{2} \\
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&=\frac{-2\pm2\sqrt{484k-44}}{2} \\
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&=\sqrt{11(11k-1)} \\
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\end{align*}</cmath>
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By the assumption that <math>n</math> is an integer, <math>11k-1</math> must has a factor of <math>11</math>, which is impossible, contradiction.
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~ Nafer
  
 
== See Also ==
 
== See Also ==
 
{{Old CanadaMO box|num-b=5|num-a=7|year=1971}}
 
{{Old CanadaMO box|num-b=5|num-a=7|year=1971}}
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]

Latest revision as of 20:18, 10 January 2020

Problem

Show that, for all integers $n$, $n^2+2n+12$ is not a multiple of $121$.

Solutions

Solution

Notice $n^{2} + 2n + 12 = (n+1)^{2} + 11$. For this expression to be equal to a multiple of 121, $(n+1)^{2} + 11$ would have to equal a number in the form $121x$. Now we have the equation $(n+1)^{2} + 11 = 121x$. Subtracting $11$ from both sides and then factoring out $11$ on the right hand side results in $(n+1)^{2} = 11(11x - 1)$. Now we can say $(n+1) = 11$ and $(n+1) = 11x - 1$. Solving the first equation results in $n=10$. Plugging in $n=10$ in the second equation and solving for $x$, $x = 12/11$. Since $12/11$ *$121$ is clearly not a multiple of 121, $n^{2} + 2n + 12$ can never be a multiple of 121.

Solution 2

Assume that $n^2+2n+12=121k$ for some integer $k$ then \[n^2+2n+(12-121k)=0\] \begin{align*} x&=\frac{-2\pm\sqrt{4-4(12-121k)}}{2} \\ &=\frac{-2\pm2\sqrt{484k-44}}{2} \\ &=\sqrt{11(11k-1)} \\ \end{align*} By the assumption that $n$ is an integer, $11k-1$ must has a factor of $11$, which is impossible, contradiction.

~ Nafer

See Also

1971 Canadian MO (Problems)
Preceded by
Problem 5
1 2 3 4 5 6 7 8 Followed by
Problem 7
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