Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 3)
(→‎Solution 3)
Line 31: Line 31:
 
reveals that only <math>10</math> satisfy the condition <math>{n^{2} + 2n + 12} \equiv 0 \pmod{11}</math>.  
 
reveals that only <math>10</math> satisfy the condition <math>{n^{2} + 2n + 12} \equiv 0 \pmod{11}</math>.  
  
−
Plugging <math>10</math> into <math>{n^{2} + 2n + 12}</math> shows that it is not divisible by 121.
+
Plugging <math>10</math> into <math>{n^{2} + 2n + 12}</math> shows that it is not divisible by <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>.
  
 
~iamselfemployed
 
~iamselfemployed

Revision as of 15:00, 15 May 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

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

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1971_Canadian_MO_Problems/Problem_6&oldid=219367"