Difference between revisions of "1971 Canadian MO Problems/Problem 6"
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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 == | + | == Solutions == |
+ | === Solution=== | ||
− | <math>n^2 + 2n + 12 = (n+1)^2 + 11</math>. | + | 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. |
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− | + | === Solution 2 === | |
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+ | 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 | ||
+ | == 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 20:18, 10 January 2020
Problem
Show that, for all integers , is not a multiple of .
Solutions
Solution
Notice . For this expression to be equal to a multiple of 121, would have to equal a number in the form . Now we have the equation . Subtracting from both sides and then factoring out on the right hand side results in . Now we can say and . Solving the first equation results in . Plugging in in the second equation and solving for , . Since * is clearly not a multiple of 121, can never be a multiple of 121.
Solution 2
Assume that for some integer then By the assumption that is an integer, must has a factor of , which is impossible, contradiction.
~ Nafer
See Also
1971 Canadian MO (Problems) | ||
Preceded by Problem 5 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • | Followed by Problem 7 |