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

Revision as of 09:45, 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

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

@@ Line 8: / Line 8: @@
 === Solution 2 ===
-Assume that <math>n^2+2n+12=121k</math> for some integer <math>k</math> then
-<cmath>n^2+2n+(12-121k)=0</cmath>
-<cmath>\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*}</cmath>
-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.
-~ Nafer
 === Solution 3 ===

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

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