The document contains solutions to two problems from an Olympiad exam. The first problem asks to find all integer pairs x and y satisfying x^2 - y^2 = 2k for some non-negative integer k, where x and y have no prime factors larger than 5. The solution lists the possible values of x^2 and y^2 and the integer pairs that satisfy the equation. The second problem asks to split a class of 12 students (6 pairs of twins) into either two teams of 6 or three teams of 4, without putting any twins in the same team. The solution explains that there are 32 ways to split them into two teams of 6, and 960 ways to split them into three

1. PROBLEM SOLVING
OLYMpiad 2005
Alfiramita Hertanti
Aulia mashar

2. 1st Question
Let x and y be positive integers with no
prime factors larger than 5. find all such
x and y wich satisfy x2-y2 =2k for some
non- negative integer k.

3. SOLUTION
Since x and y are positive integers with no
prime factors large than 5, we can express
them as follows x= 2a x 3b x 5c, and y= 2d x 3e x
5f where all the values a,b,c,d,e and f take on
the values of either 0 or 1 or 2, and the
possible values for x2 and y2 are
X2 = 1,4,9,25,36,100,225,900.
Y2 = 1,4,9,25,36,100,225,900.

4. The problem requires x>y and the difference of x2-y2
to be an even number 0r 1. and (k=0,1,2,3…..)
Therefore,
(x2,y2) =
(9,1), (25,1), (225,1), (25,9), (225,9), (225,25), (36,4), (100
,4), (900,4), (100,36), (900,36),
(900,100).
Which meets 2k
(x2,y2) = (9,1), (25,9), (36,4), (100,36)
(x,y) = (3,1), (5,3), (6,2) , (10,6)

5. 2nd Question
Adrian teaches a class of six pairs of twins. He wishes
to set up teams for a quiz, but wants to avoid putting
any pair of twins into the same team. Subject to this
condition:
In how many ways can he split them into two teams of
six?
In how many ways can he split them into three teams
of four?

6. i)
Team 1 must contain one person from each of the
pairs of twins. There’s 2 ways in which we can pick
a person from each pair, so 26 ways in total of
picking people for Team 1. But we have to consider
duplicates – had all the people in Team 1 actually
been in Team 2 instead, it would have been
considered the same arrangement (i.e. the teams
themselves are indistinguishable even if the people
in each team are obviously distinguishable). So
that gives 25 = 32 ways.

7. There’s
ways of choosing the pairs for which
we’re going to use 1 person from each to form Team 1.
From each pair we can choose one twin, so that’s 15 x
24 = 240 ways. For Team 2, we have four
individuals, and two pairs left to choose from. But we
have to be careful now, because if we picked say the 4
individuals to form the next team, we’d have two pairs
of twins left to form the final team, which is not
permitted. We have to ensure therefore that one
person from each of the remaining intact pairs is
chosen for Team 2 (22 ways). There’s then
ways
in which we can choose from the individuals to fill up
the rest of the team. That gives 22 x 6 = 24 ways.
Team 3 is then fixed because we have 4 people left.
So in total we 240 x 24 = 5760 ways. But again, the
teams are ‘indistinguishable’, so we have to divide by 3!
to account for the possible relabeling of teams. That
gives 960 ways.
ii)