Difference between revisions of "1971 Canadian MO Problems/Problem 6"
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~ Nafer | ~ Nafer | ||
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+ | === 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 none of these numbers satisfy the condition <math>{n^{2} + 2n + 12} \equiv 0 \pmod{11}</math>. | ||
+ | |||
+ | Thus, there is no integer <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]] |
Revision as of 14:53, 15 May 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
Assume that for some integer
then
By the assumption that
is an integer,
must has a factor of
, which is impossible, contradiction.
~ Nafer
Solution 3
In order for to divide
,
must also divide
.
Plugging in all numbers modulo :
or
to make computations easier,
reveals that none of these numbers satisfy the condition .
Thus, there is no integer 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 |