CSAT For Computer Science Undergrad Admissions 2016 Entry

Consider the definition of an irrational number. Can you re-arrange the equation to arrive at a condradiction? (Hint: Making b the subject probably won't get you very far.)

So terrified for this exam. Especially since I'm struggling on Q1 xD Can't get anything other than b = a(sqrt(2)) - c(sqrt(3)) and since sqrt(2) and sqrt(3) are irrational, integer multiples of them are still irrational, so the difference between integer multiples of them cannot be a positive integer. (Like you can't make npi = a(sqrt(2)), etc.). So it's a,b,c=0? But, doesn't feel right xD
I seem to be working it out massively different to everyone else for Q2?
EDIT: Screw that question about the earths surface, absolutely no clue.

Consider the definition of an irrational number. Can you re-arrange the equation to arrive at a condradiction? (Hint: Making b the subject probably won't get you very far.)

You only have to be able to answer half of the Section A questions at best, so if you can do 1 or 2 of the 4 practice ones then you should be fine.

I've answered your spoiler question in this spoiler

Like, I don't know if I should answer 7 section A's then just move on to section B? Or should I answer 8 section A's incase I got one wrong?

The paper rubric clearly states to answer no more than 7 questions in Section A. So I don't think it would make a good impression if you deliberately answered more than that.

I'm not a CompSci applicant, but I am a maths one, so I'll try my hand at a few of these questions, or just pick a few and do them in several ways. :-)

Q1:

Method 1:
Method 2:
Q25:

Method 1:
Method 2:

FLT:

Thanks for making this! I need some general tips though. For question 25, method 1.

You seemed to have just known to rewrite that last bracket in that form... So that you got the product of 5 consecutive integers. How on earth did you just decide to do this? I would never even think of doing that?!?! How can I make these amazing decisions? xD

Because in n consecutive integers , at least one of them is divisible by n. Since he knew this, he approached the question wanting to use that property. That is one of the most common way to prove that an expression is divisible by n - writing it as product of n cons. integers.

If you do enough questions, you will eventually see a similar question and you would know what to do. And if you see a unfamiliar problem? One of the biggest tricks in academics is to make an unfamiliar problem into a familiar problem and then you can solve it immediately. But that comes with practice. For me, I participate in national team training (hue,hue) never made the team tho.

Thanks for making this! I need some general tips though. For question 25, method 1.

You seemed to have just known to rewrite that last bracket in that form... So that you got the product of 5 consecutive integers. How on earth did you just decide to do this? I would never even think of doing that?!?! How can I make these amazing decisions? xD

A product of $r$ consecutive integers is divisible by $r!$.

............... That is kind of obvious ==" by decreasing values of r.

Because in n consecutive integers , at least one of them is divisible by n. Since he knew this, he approached the question wanting to use that property. That is one of the most common way to prove that an expression is divisible by n - writing it as product of n cons. integers.

If you do enough questions, you will eventually see a similar question and you would know what to do. And if you see a unfamiliar problem? One of the biggest tricks in academics is to make an unfamiliar problem into a familiar problem and then you can solve it immediately. But that comes with practice. For me, I participate in national team training (hue,hue) never made the team tho.

A product of $r$ consecutive integers is divisible by $r!$.

............... That is kind of obvious ==" by decreasing values of r.

... I know it's obvious. Just like what you said was obvious...

Are you basically saying that it's a general technique when checking if an algebraic expression is divisible by a number n? Find a way to express it into only multiples of consecutive (or single) integers, so that by using the factors of n we can try to find a way to factorise that expression so that it gives n x something?





2 Maths genuines basically just calling me stupid

Well, no, not really. A general technique when given a polynomial-like algebraic expression is to factorise it.

And no, of course not! You're not stupid at all.

Yeah I understand the factorising idea, but seeing you turn (n^2 + 1) into (n^2 - 4 + 5) then just deciding to take the 5 out was just what everyone just skips over? Like, you just did it, I wouldn't even think of doing that xD Hopefully I'll learn to do this more often - rewrite the bracket in a seemingly random way, but actually do it strategically in order to help me in a later step. Thinking about this, it almost requires you to know what you're doing backwards, so you can specifically search for a way to put it in this particular form.

Haha I'll take that, you're definitely one of the better mathematicians I see on these forums

Sometimes you kinda just need to rewrite things the way you want them
For example: let's say you want to integrate $\displaystyle \int \frac{x}{x+1} \, \mathrm{d}x$ it doesn't look so easy, but if you just add 0 to the numerator (in a specific way), you get

$\displaystyle \frac{x+1 - 1}{x+1} = \frac{x+1}{x+1} - \frac{1}{x+1} = 1 - \frac{1}{x+1}$

which is very easy to integrate. You get $x - \ln |x+1| + c$

That's very kind of you.