Difference between revisions of "2013 AMC 10B Problems/Problem 22"
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Isabelchen (talk | contribs) (→Side Note to Solution 1 & 2) |
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<math>4S=45+3J</math> | <math>4S=45+3J</math> | ||
<math>4S=3(15+J)</math> | <math>4S=3(15+J)</math> | ||
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Because <math>(4,3)=1</math>, so <math>4|(15+J)</math>, but <math>15 \equiv 3 \pmod 4</math>, so <math>J \equiv 1 \pmod 4</math> | Because <math>(4,3)=1</math>, so <math>4|(15+J)</math>, but <math>15 \equiv 3 \pmod 4</math>, so <math>J \equiv 1 \pmod 4</math> | ||
So <math>J=1,5,9</math> | So <math>J=1,5,9</math> | ||
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When <math>J=1, S=12</math>, <math>A+E=B+F=C+G=D+H=11</math> | When <math>J=1, S=12</math>, <math>A+E=B+F=C+G=D+H=11</math> | ||
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When <math>J=5, S=15</math>, <math>A+E=B+F=C+G=D+H=10</math> | When <math>J=5, S=15</math>, <math>A+E=B+F=C+G=D+H=10</math> | ||
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When <math>J=9, S=18</math>, <math>A+E=B+F=C+G=D+H=9</math> | When <math>J=9, S=18</math>, <math>A+E=B+F=C+G=D+H=9</math> | ||
~isabelchen | ~isabelchen | ||
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==Video Solution by TheBeautyofMath== | ==Video Solution by TheBeautyofMath== |
Revision as of 11:34, 28 September 2021
Contents
Problem
The regular octagon has its center at
. Each of the vertices and the center are to be associated with one of the digits
through
, with each digit used once, in such a way that the sums of the numbers on the lines
,
,
, and
are all equal. In how many ways can this be done?
Solution 1
First of all, note that must be
,
, or
to preserve symmetry, since the sum of 1 to 9 is 45, and we need the remaining 8 to be divisible by 4 (otherwise we will have uneven sums). So, we have:
We also notice that .
WLOG, assume that . Thus the pairs of vertices must be
and
,
and
,
and
, and
and
. There are
ways to assign these to the vertices. Furthermore, there are
ways to switch them (i.e. do
instead of
).
Thus, there are ways for each possible J value. There are
possible J values that still preserve symmetry:
Solution 2
As in solution 1, must be
,
, or
giving us 3 choices. Additionally
. This means once we choose
there are
remaining choices. Going clockwise from
we count,
possibilities for
. Choosing
also determines
which leaves
choices for
, once
is chosen it also determines
leaving
choices for
. Once
is chosen it determines
leaving
choices for
. Choosing
determines
, exhausting the numbers. Additionally, there are three possible values for
. To get the answer we multiply
.
Side Note to Solution 1 & 2
Solution 1 and 2 states that without rigor analysis. Here it is:
Let![]()
![]()
![]()
![]()
Because , so
, but
, so
So
When,
When
,
When
,
![]()
~isabelchen
Video Solution by TheBeautyofMath
~IceMatrix
See also
2013 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.