- This is an exam question
given this year to high school students in Turkey.
A project team consists of a hundred people.
Every member is appointed to an equal number of projects.
No two people are appointed to the exact same projects.
The situation cannot be achieved if every member
is appointed to three projects, but can be achieved
if every member is appointed to four projects.
So how many projects does the team have?
six, seven, eight, nine, or 10.
Have a think, I'll answer in a minute,
but first let's take a step back and think about
how we got here.
This question is from an exam paper called AYT,
which has one of two papers given to high school students
in Turkey who wished to gain entrance to university.
It is sat by over 2.5 million students per year.
And contains questions about Turkish language,
social science, mathematics, and science.
And as this a common theme with exams like this
from around the world,
this exam can really be a make or break moment
in the lives of young people.
Turkey has its own breed of cram schools,
and I've received plenty of comments from Turkish students
about this exam and how it has impacted their life.
Here's one that says, of the more than 2 million people
who take it every year, that only a few rank high enough
to get into the best universities.
It is considered as an only chance
at a good future by everyone.
Most people start studying two years before the exam.
And in the last year,
people can study up to 10 to 15 plus hours a day.
You'd better hope that you're a good all-rounder
because the exam paper combines
all of the different subjects together,
and you have 118 minutes to answer 166 questions.
This math question I showed was number 15 on the 2021 paper,
and was particularly controversial
because of its difficulty.
A similar type of problem appeared
on the Turkish Mathematical Olympiad.
So students were left wondering
why an Olympiad level question was given to everyone,
including those who have no intention to major in math.
To understand it, I wanna talk about something
called the pigeonhole principle.
In mathematics the pigeonhole principle states that
if n items I put into m containers
with n being greater than m,
then at least one container must contain more than one item.
So over here, if we have 10 pigeons dispersed within
nine pigeon holes, then since 10 is greater than nine,
then there will be at least one pigeonhole
that has more than one pigeon.
And up here, we can see this one has two pigeons in it.
Honestly, a little too obvious,
a little too basic to seem useful,
but actually we can use it to solve our problem.
Our problem is one of finding the containers.
Our items n is 100 people and these hundred people
are being dispersed across the containers,
but our containers are not the projects.
Our containers are the unique sets of projects
that each person is appointed to.
So if x is the total number of projects,
then the set of all projects goes from P1 up to Px.
If everyone is appointed to three projects,
then the containers would represent each unique combination
of three projects.
The first container might have P1, P2, P3,
the next one P1, P2, P4 then P1, P2, P5
and so on until we've had every unique combination.
The question requires that no two people are appointed
to the exact same projects, which would be two people,
or I guess we can use turkeys in one container.
But we also know from the question that
for three projects, each, the situation doesn't work out.
So actually we are going to have at least one container
with more than one person in it.
So let's find out when that happens.
According to the pigeonhole principle,
it's when m is greater than m.
So when we have more people or turkeys
than we do containers.
We know that we have a hundred people,
so if the number of containers is anything
less than a hundred, we're going to have some double-ups.
But how do we find is the number of containers?
How do we find the unique combinations of three projects?
Well, we can use the combinations formula n choose r.
also it can be written like this.
So we want x choose three to be less than a hundred.
We know we're going to have some double-ups.
Our options for x are six, seven, eight, nine or 10.
And of these 10 choose three is the only one
that's not less than a hundred,
so it can be ruled out as a possible answer.
Now, let's see about four projects.
Each we know that this time it will all work out
and there will be no double ups.
Our containers are all the unique combinations
of four projects.
And the number of containers is given by x choose four.
And this time we want that value to be more than
or equal to 100, six, seven, and eight choose four
all their value is less than a hundred.
So they are ruled out as possible options.
And the only remaining option for X
that satisfies both criteria is nine.
So we can say that the correct answer here is D.
Here's that question,
as it was originally presented in Turkish.
We can look through and see some of the other questions.
For this one you have P of X, a fourth degree polynomial.
You're given all the conditions you need to know about P
and then asked to find what is P at four.
In 17, you're asked to find the limit
of a rather nasty function without using a calculator.
Moving on to some physics.
This question here asks us about the relationship between
the velocities of three different paths of the golf ball.
If we go back to maths for a second,
we can find this question here,
which can be translated into English
to say that m and n are positive integers,
the GCD of m and n plus the LCM of m and n is 289,
m plus n is not equal to 289.
So find the total m plus n.
This one is near the start of the paper
and seems like it might be a bit easier,
but actually it can be a bit tricky.
To help us understand this one,
I'm going to turn to Brilliant,
who has sponsored the section of the video.
We have the situation that hotdogs
are sold in packages of 10,
while the buns are sold in packages of eight.
Suppose you want to buy exactly the same number of hot dogs
and buns, What is the minimum amount you will need to buy?
This is us having to find
the lowest common multiple, the LCM.
These are the multiples of 10.
These are the multiples of eight.
And the lowest common one is 40.
Next question, if we wanted to reduce the fraction
24/30 to lowest terms, what number should you divide
the numerator and denominator by?
The answer is six, and that is our, GCD,
the greatest common divisor.
These are all the numbers 24 can be divided by.
These are all the numbers that 30 can be divided by.
And the greatest common one is six.
Going back to our question,
let's just call the GCD of m and n sum value x.
That would mean m can be written as x times sum value a,
and in would be x times sum value b.
Now a and b are relatively prime.
There's no common factor between them
and nothing further that you could divide them both by.
That would mean that the lowest common multiple
of m and n is x times a times b.
So our equation up here becomes x plus x times ab is 289.
We can rewrite that as x times one plus ab.
Here in the table of possible values
that would make this true.
We can already rule out the second option
because if one plus ab was equal to one,
then ab would have to be equal to zero.
And we know that's not true
because m and n are positive integers.
So we can try the first option x is one,
and one plus ab is 289.
So ab is 288.
Being able to factor as this into its prime factors
and therefore being able to work out what possible
factors it has is a crunch point for this question.
Then in this little table here,
I've put the two options for either a or b.
Looking at the total m plus n for the first option,
we would end up with one plus 288, which is 289.
However, the question says that that is not allowed.
So this is not a possible option.
Trying the next option, we have 32 plus nine, which is 41.
That's probably the right answer.
But we still had one option to try over here,
which is when x and one plus ab both equal to 17,
'cause 17 squared is 289.
To get eight times b equal to 16,
one has to be one, and the other has to be 16.
You might think that two and eight are also options,
but if you remember, a and B are relatively prime
and two and eight have a common factor of two.
So aren't relatively prime.
This option also ends up giving us that m plus n is 289,
which we can't have.
So the correct answer to this question is indeed 41.
You can do more LCM and GCD problems on Brilliant,
Head to brilliant.org/tibees to try for yourself.
Thanks Brilliant for supporting this video.
And also thanks to my Patreon supporters.
A special shout out to today's Patron cat of the day, Ember.
(soft music)