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

(Solution)
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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 1 ===
 
<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>.
 
<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>.
 
The quadratic residues <math>mod 11</math> are <math>1, 3, 4, 5, 9</math>, and <math>0</math> (as shown below).
 
The quadratic residues <math>mod 11</math> are <math>1, 3, 4, 5, 9</math>, and <math>0</math> (as shown below).
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Thus, for any integer <math>n</math>, <math>n^2+2n+12</math> is not a multiple of <math>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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=== Solution 2 ===
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<math>n^2+2n+12=(n+1)^2+11</math>, so if this is a multiple of 11 then <math>(n+1)^2</math> is too. Since 11 is prime, <math>n+1</math> must be divisible by 11, and hence <math>(n+1)^2</math> is divisible by 121. This shows that <math>(n+1)^2+11</math> can never be divisible by 121.
  
  
 
{{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]]

Revision as of 15:16, 12 September 2012

Problem

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

Solutions

Solution 1

$n^2 + 2n + 12 = (n+1)^2 + 11$. Consider this equation mod 11. $(n+1)^2 + 11 \equiv (n+1)^2 \mod 11$. The quadratic residues $mod 11$ are $1, 3, 4, 5, 9$, and $0$ (as shown below).

If $n \equiv 0 \mod 11$, $(n+1)^2 \equiv (0+1)^2 \equiv 1\mod 11$, thus not a multiple of 11, nor 121.

If $n \equiv 1 \mod 11$, $(n+1)^2 \equiv (1+1)^2 \equiv 4\mod 11$, thus not a multiple of 11, nor 121.

If $n \equiv 2 \mod 11$, $(n+1)^2 \equiv (2+1)^2 \equiv 9\mod 11$, thus not a multiple of 11, nor 121.

If $n \equiv 3 \mod 11$, $(n+1)^2 \equiv (3+1)^2 \equiv 5\mod 11$, thus not a multiple of 11, nor 121.

If $n \equiv 4 \mod 11$, $(n+1)^2 \equiv (4+1)^2 \equiv 3\mod 11$, thus not a multiple of 11, nor 121.

If $n \equiv 5 \mod 11$, $(n+1)^2 \equiv (5+1)^2 \equiv 3\mod 11$, thus not a multiple of 11, nor 121.

If $n \equiv 6 \mod 11$, $(n+1)^2 \equiv (6+1)^2 \equiv 5\mod 11$, thus not a multiple of 11, nor 121.

If $n \equiv 7 \mod 11$, $(n+1)^2 \equiv (7+1)^2 \equiv 9\mod 11$, thus not a multiple of 11, nor 121.

If $n \equiv 8 \mod 11$, $(n+1)^2 \equiv (8+1)^2 \equiv 4\mod 11$, thus not a multiple of 11, nor 121.

If $n \equiv 9 \mod 11$, $(n+1)^2 \equiv (9+1)^2 \equiv 1\mod 11$, thus not a multiple of 11, nor 121.


If $n \equiv 10 \mod 11$, $(n+1)^2 \equiv (10+1)^2 \equiv 0\mod 11$. However, considering the equation $\mod 121$ for $n \equiv 10 \mod 11$, testing $n = 10, 21, 32, 43, 54, 65, 76, 87, 98, 109, 120$, we see that $(n+1)^2 + 11$ always leave a remainder of greater than $1\mod 121$.

Thus, for any integer $n$, $n^2+2n+12$ is not a multiple of $121$.

Solution 2

$n^2+2n+12=(n+1)^2+11$, so if this is a multiple of 11 then $(n+1)^2$ is too. Since 11 is prime, $n+1$ must be divisible by 11, and hence $(n+1)^2$ is divisible by 121. This shows that $(n+1)^2+11$ can never be divisible by 121.


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