Difference between revisions of "1971 Canadian MO Problems/Problem 6"
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== Solutions == | == Solutions == | ||
− | === Solution=== | + | === Solution 1 === |
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. | 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. | ||
=== Solution 2 === | === 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 <math>121</math> to divide <math>n^{2} + 2n + 12</math>, <math>11</math> must also divide <math>n^{2} + 2n + 12</math>. | ||
+ | |||
+ | Plugging in all numbers modulo <math>11</math>: | ||
+ | |||
+ | <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, | ||
+ | |||
+ | 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 <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 | ||
== 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 , is not a multiple of .
Solutions
Solution 1
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
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 to divide , must also divide .
Plugging in all numbers modulo :
or to make computations easier,
reveals that only satisfy the condition .
Plugging into shows that it is not divisible by .
Thus, there are no integers such that is divisible by .
~iamselfemployed
See Also
1971 Canadian MO (Problems) | ||
Preceded by Problem 5 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • | Followed by Problem 7 |