Problem
of the Week
Problem
E and Solution
Take
a Seat 3
Problem
Twelve people are sitting, equally spaced, around a circular table.
They each hold a card with a different integer on it. For any two people
sitting beside each other, the positive difference between the integers
on their cards is no more than .
The people holding the integers
and are seated as shown. The
person opposite the person holding the is holding the integer . What are the possible values of ?
Solution
We will start with the card with the integer . We are given that is on one side of the . Let be the integer on the other side of the
.
Since each card contains a different integer and the positive
difference between the integers on two cards beside each other is no
more than , then must be , , or
. We will consider these three
cases.
Case 1:
Since the number on each card is different and we know that someone
is holding a card with a and
someone is holding a card with a ,
then the integer to the right of the must be or . Furthermore, every integer to the
right of must be greater than
. Similarly, the integer to left
of is either or . Furthermore, any integer to the left
of must be less than .
Since is both to the right of
and to the left of , it must be both greater than and less than . This is not possible.
Therefore, when , there is no
solution for .
Case 2:
Since the number on each card is different and we know that someone
is holding a card with a and
someone is holding a card with a ,
then the integer to the right of
must be or . We will look at these two
subcases.
Case 2a: The card with integer is to the right of the card with
integer .
Notice then that every integer to the right of the must be less than . Also the integer to the left of must be and every integer to left of the must be greater than . Since is both to the right of and to the left of , it must be both greater than and less than . This is not possible.
Therefore, when and the
integer to the right of it is , there is no
solution for .
Case 2b: The card with integer is to the right of the card with
integer .
Now, the integer to the left of can be either or .
If the integer is , then using
a similar argument to that in Case 2a, there is no solution for .
If the integer to the left of is
, the only possible integer to the
left of is . This means the only possible integer
to the right of is . Which leads to the only possible
integer to the left of is . Furthermore, the only possible
integer to the right of is . Continuing in this manner, we get the
table set up shown below.
From here, the only possible solution is .
Therefore, when , the
solution is .
Case 3:
Since the number on each card is different and we know that someone
is holding a card with a , then
the integer to right of the must
be , , or . Furthermore, since someone is already
holding a and someone is already
holding a , every other integer to
the the right of must be or greater.
The integer to the left of is
either , , or . If it is or , then since someone is already holding
the and someone is already
holding the , every integer to the
left of must be less than . Since is both to the right of and to the left of , if there is a or a to the left of , then must be both or greater and less than . This is not possible.
Therefore, if a solution exists when , then the integer to the left of
must be
. The integer to the left of must be , , , or . Since the , , and are already placed, then the only
possible integer to the left of
is . Similarly, the only possible
integer to the right of is . Thus, the integer to the left of
must be . Continuing in this manner, we get the
table set up shown below.
From here, the only possible solution is .
Therefore, when , the
solution is .
Therefore, the possible values for are or .