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

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== Solutions ==
 
== Solutions ==
 
=== Solution 1 ===
 
=== Solution 1 ===
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<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).
 
  
−
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.  
+
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.
  
−
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.
+
=== Solution 2 ===
−
 
+
n^2+2n+12
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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.
+
=(n+1)^2+11
−
 
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= (n+1)(n+1)+11
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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.
+
Now 11 itself is a multiple of 11, Therefore there are 2 cases for the value of n
 +
Case 1: n+1 isn't a multiple of 11
 +
if n isn't a multiple of 11 then (n+1)^2 isn't a multiple of 121 i.e n^2+2n+12 Isn't divisible by 11
 +
Case 2: if n+1 is a multiple of 11
 +
then (n+1)^2 is a multiple of 121, let (n+1)^2=121k
 +
But 121k+11 can't be equal to a multiple of 11
 +
hence proved
  
−
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.
+
=== Solution 3 ===
  
−
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.
+
In order for <math>121</math> to divide <math>n^{2} + 2n + 12</math>, <math>11</math> must also divide <math>n^{2} + 2n + 12</math>.  
−
 
−
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.
+
Plugging in all numbers modulo <math>11</math>:
  
−
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.
+
<math>0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10,</math>
 +
or <math>0, 1, 2, 3, 4, 5, (-5), (-4), (-3), (-2), (-1)</math> to make computations easier,  
  
−
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.
+
reveals that only <math>10</math> satisfy the condition <math>{n^{2} + 2n + 12} \equiv 0 \pmod{11}</math>.  
−
 
  
−
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>.  
+
Plugging <math>10</math> into <math>{n^{2} + 2n + 12}</math> shows that it is not divisible by <math>121</math>.
  
−
Thus, for any integer <math>n</math>, <math>n^2+2n+12</math> is not a multiple of <math>121</math>.
+
Thus, there are no integers <math>n</math> such that <math>n^{2} + 2n + 12</math> is divisible by <math>121</math>.
  
−
=== Solution 2 ===
+
~iamselfemployed
−
<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.
 
  
 
== 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 19:05, 12 June 2024

Problem

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

Solutions

Solution 1

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

n^2+2n+12 =(n+1)^2+11 = (n+1)(n+1)+11 Now 11 itself is a multiple of 11, Therefore there are 2 cases for the value of n Case 1: n+1 isn't a multiple of 11 if n isn't a multiple of 11 then (n+1)^2 isn't a multiple of 121 i.e n^2+2n+12 Isn't divisible by 11 Case 2: if n+1 is a multiple of 11 then (n+1)^2 is a multiple of 121, let (n+1)^2=121k But 121k+11 can't be equal to a multiple of 11 hence proved

Solution 3

In order for $121$ to divide $n^{2} + 2n + 12$, $11$ must also divide $n^{2} + 2n + 12$.

Plugging in all numbers modulo $11$:

$0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10,$ or $0, 1, 2, 3, 4, 5, (-5), (-4), (-3), (-2), (-1)$ to make computations easier,

reveals that only $10$ satisfy the condition ${n^{2} + 2n + 12} \equiv 0 \pmod{11}$.

Plugging $10$ into ${n^{2} + 2n + 12}$ shows that it is not divisible by $121$.

Thus, there are no integers $n$ such that $n^{2} + 2n + 12$ is divisible by $121$.

~iamselfemployed

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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