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

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(Solution 2)
 
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Show that, for all integers <math>n</math>, <math>n^2+2n+12</math> is not a multiple of <math>121</math>.  
 
Show that, for all integers <math>n</math>, <math>n^2+2n+12</math> is not a multiple of <math>121</math>.  
  
== Solution ==
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== Solutions ==
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=== Solution===
  
<math>n^2 + 2n + 12 = (n+1)^2 + 11</math>. Consider this equation mod 11. <math> (n+1)^2 + 11 \equiv (n+1)^2 \mod 11</math>.
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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.
The quadratic residues <math>mod 11</math> are <math>1, 3, 4, 5, 9</math>, and <math>0</math> (as shown below).
 
  
If <math>n \equiv 0 \mod 11</math>, <math>(n+1)^2 \equiv (0+1)^2 \equiv 1\mod 11</math>, thus not a multiple of 11, nor 121.
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=== Solution 2 ===
 
 
If <math>n \equiv 1 \mod 11</math>, <math>(n+1)^2 \equiv (1+1)^2 \equiv 4\mod 11</math>, thus not a multiple of 11, nor 121.
 
 
 
If <math>n \equiv 2 \mod 11</math>, <math>(n+1)^2 \equiv (2+1)^2 \equiv 9\mod 11</math>, thus not a multiple of 11, nor 121.
 
 
 
If <math>n \equiv 3 \mod 11</math>, <math>(n+1)^2 \equiv (3+1)^2 \equiv 5\mod 11</math>, thus not a multiple of 11, nor 121.
 
 
 
If <math>n \equiv 4 \mod 11</math>, <math>(n+1)^2 \equiv (4+1)^2 \equiv 3\mod 11</math>, thus not a multiple of 11, nor 121.
 
 
 
If <math>n \equiv 5 \mod 11</math>, <math>(n+1)^2 \equiv (5+1)^2 \equiv 3\mod 11</math>, thus not a multiple of 11, nor 121.
 
 
If <math>n \equiv 6 \mod 11</math>, <math>(n+1)^2 \equiv (6+1)^2 \equiv 5\mod 11</math>, thus not a multiple of 11, nor 121.
 
 
 
If <math>n \equiv 7 \mod 11</math>, <math>(n+1)^2 \equiv (7+1)^2 \equiv 9\mod 11</math>, thus not a multiple of 11, nor 121.
 
 
 
If <math>n \equiv 8 \mod 11</math>, <math>(n+1)^2 \equiv (8+1)^2 \equiv 4\mod 11</math>, thus not a multiple of 11, nor 121.
 
 
 
If <math>n \equiv 9 \mod 11</math>, <math>(n+1)^2 \equiv (9+1)^2 \equiv 1\mod 11</math>, thus not a multiple of 11, nor 121.
 
 
 
 
If <math>n \equiv 10 \mod 11</math>,  <math>(n+1)^2 \equiv (10+1)^2 \equiv 0\mod 11</math>.  However, considering the equation <math>\mod 121</math> for <math>n \equiv 10 \mod 11</math>, testing <math>n = 10, 21, 32, 43, 54, 65, 76, 87, 98, 109, 120</math>, we see that <math>(n+1)^2 + 11</math> always leave a remainder of greater than <math>1\mod 121</math>.
 
 
 
Thus, for any integer <math>n</math>, <math>n^2+2n+12</math> is not a multiple of <math>121</math>.
 
  
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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
  
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== 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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