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

Revision as of 14:16, 12 September 2012

Problem

Show that, for all integers $n$ , $n^2+2n+12$ is not a multiple of $121$ .

Solutions

Solution 1

$n^2 + 2n + 12 = (n+1)^2 + 11$ . Consider this equation mod 11. $(n+1)^2 + 11 \equiv (n+1)^2 \mod 11$ . The quadratic residues $mod 11$ are $1, 3, 4, 5, 9$ , and $0$ (as shown below).

If $n \equiv 0 \mod 11$ , $(n+1)^2 \equiv (0+1)^2 \equiv 1\mod 11$ , thus not a multiple of 11, nor 121.

If $n \equiv 1 \mod 11$ , $(n+1)^2 \equiv (1+1)^2 \equiv 4\mod 11$ , thus not a multiple of 11, nor 121.

If $n \equiv 2 \mod 11$ , $(n+1)^2 \equiv (2+1)^2 \equiv 9\mod 11$ , thus not a multiple of 11, nor 121.

If $n \equiv 3 \mod 11$ , $(n+1)^2 \equiv (3+1)^2 \equiv 5\mod 11$ , thus not a multiple of 11, nor 121.

If $n \equiv 4 \mod 11$ , $(n+1)^2 \equiv (4+1)^2 \equiv 3\mod 11$ , thus not a multiple of 11, nor 121.

If $n \equiv 5 \mod 11$ , $(n+1)^2 \equiv (5+1)^2 \equiv 3\mod 11$ , thus not a multiple of 11, nor 121.

If $n \equiv 6 \mod 11$ , $(n+1)^2 \equiv (6+1)^2 \equiv 5\mod 11$ , thus not a multiple of 11, nor 121.

If $n \equiv 7 \mod 11$ , $(n+1)^2 \equiv (7+1)^2 \equiv 9\mod 11$ , thus not a multiple of 11, nor 121.

If $n \equiv 8 \mod 11$ , $(n+1)^2 \equiv (8+1)^2 \equiv 4\mod 11$ , thus not a multiple of 11, nor 121.

If $n \equiv 9 \mod 11$ , $(n+1)^2 \equiv (9+1)^2 \equiv 1\mod 11$ , thus not a multiple of 11, nor 121.

If $n \equiv 10 \mod 11$ , $(n+1)^2 \equiv (10+1)^2 \equiv 0\mod 11$ . However, considering the equation $\mod 121$ for $n \equiv 10 \mod 11$ , testing $n = 10, 21, 32, 43, 54, 65, 76, 87, 98, 109, 120$ , we see that $(n+1)^2 + 11$ always leave a remainder of greater than $1\mod 121$ .

Thus, for any integer $n$ , $n^2+2n+12$ is not a multiple of $121$ .

Solution 2

$n^2+2n+12=(n+1)^2+11$ , so if this is a multiple of 11 then $(n+1)^2$ is too. Since 11 is prime, $n+1$ must be divisible by 11, and hence $(n+1)^2$ is divisible by 121. This shows that $(n+1)^2+11$ can never be divisible by 121.

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

@@ Line 2: / Line 2: @@
 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 1 ===
 <math>n^2 + 2n + 12 = (n+1)^2 + 11</math>. Consider this equation mod 11. <math> (n+1)^2 + 11 \equiv (n+1)^2 \mod 11</math>.
 The quadratic residues <math>mod 11</math> are <math>1, 3, 4, 5, 9</math>, and <math>0</math> (as shown below).
@@ Line 32: / Line 32: @@
 Thus, for any integer <math>n</math>, <math>n^2+2n+12</math> is not a multiple of <math>121</math>.
+=== Solution 2 ===
+<math>n^2+2n+12=(n+1)^2+11</math>, so if this is a multiple of 11 then <math>(n+1)^2</math> is too. Since 11 is prime, <math>n+1</math> must be divisible by 11, and hence <math>(n+1)^2</math> is divisible by 121. This shows that <math>(n+1)^2+11</math> can never be divisible by 121.
 {{Old CanadaMO box|num-b=5|num-a=7|year=1971}}
 [[Category:Intermediate Algebra Problems]]

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