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

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=== Solution 2 ===
 
=== Solution 2 ===
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n^2+2n+12
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Assume that <math>n^2+2n+12=121k</math> for some integer <math>k</math> then
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=(n+1)^2+11
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<cmath>n^2+2n+(12-121k)=0</cmath>
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= (n+1)(n+1)+11
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<cmath>\begin{align*}
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Now 11 itself is a multiple of 11, Therefore there are 2 cases for the value of n
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x&=\frac{-2\pm\sqrt{4-4(12-121k)}}{2} \\
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Case 1: n+1 isn't a multiple of 11
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&=\frac{-2\pm2\sqrt{484k-44}}{2} \\
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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
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&=\sqrt{11(11k-1)} \\
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Case 2: if n+1 is a multiple of 11
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\end{align*}</cmath>
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then (n+1)^2 is a multiple of 121, let (n+1)^2=121k
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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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But 121k+11 can't be equal to a multiple of 11
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hence proved
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~ Nafer
 
  
 
=== Solution 3 ===
 
=== Solution 3 ===

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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