Oxbridge Maths Interview Questions - Daily Rep

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They asked me about staircases...somehow!!! (sorry, bit of on 'in joke')

Similar stuff I guess. Am looking after interviewees on 7th/8th up at John's, gotta keep hush hush 'n' all. :p:

Reply 21

dvs

vector771

Does anyone know how to prove Pascal's triangle?

C(n,k) = C(n-1, k) + C(n-1, k-1)

(Algebra.)

Reply 22

Kolya

Q75. Prove for a^2 + b^2 = c^2 a and b can't both be odd.

Let us assume a and b both are odd, and write them as (2n+1) and (2k+1) respectively, where n and k are integers.
(2n+1)^2 + (2k+1)^2 = c^2
4n^2 + 4n + 1 + 4k^2 + 4k + 1 = c^2
2(2n^2 + 2n + 2k^2 + 2k + 1) = c^2
But, whatever the bracket is, when we square root we get a multiple of the square root of 2, which is irrational, therefore a and b cannot both be odd.

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Q58. What are the possible unit digits for perfect squares?
1,4,9,6,5,0
(Taken from squares of nos. 1-10, any greater have units digit which is square of its unit digit, or a nos. 1-10.)

Reply 23

vector

Consider 1*2*3*4*5*6*... Every fifth number, starting with 5, is divisible by 5. That gives 100/5 = 20 factors of 5 in 100!. Also, every 25th number, starting with 25, has an extra factor of 5 beyond the ones already counted. That gives you 100/25 = 4 more factors of 5 in 100!. To get a third factor of 5 from a single number, it has to be a multiple of 125, and no number <= 100 is, so that is all.

The answer is: [100/5] + [100/5^2] + [100/5^3] + ... = 20 + 4 + 0 + ... = 24

Reply 24

Dystopia

Lusus Naturae

Q58. What are the possible unit digits for perfect squares?
1,4,9,5,0
(Taken from squares of nos. 1-10, any greater have units digit which is square of its unit digit, or a nos. 1-10.)

Why isn't 6 included? :s-smilie:

Reply 25

Kolya

Dystopia

Why isn't 6 included? :s-smilie:

Because I cannot do Maths.

Reply 26

vector

Lusus Naturae

Because I cannot do Maths.

You mean your 6 key is broken :smile:

Reply 27

zrancis

99) let four consective numbers = n-1, n, n+1, n+2

For products of n-1, n one of these must be even, therefore it is divisible by 2

For products of n-1, n, n+1, one of these must be even, and one of these must be a multiple of 3, therefore it is divisible by 6

For products of n-1, n, n+1, n+2, two must be even, one a multiple of 2 and the other a multiple of 4, therefore it must be divisible by 8 and one of these must be a multiple of 3 therefore it must be a multiple of 24

Reply 28

insparato

21. Integrate and differentiate xln(x)

By parts

Unparseable latex formula:

\large \tex \int xlnx

u = ln x

Unparseable latex formula:

\large \tex \frac{du}{dx} = \frac{1}{x}

Unparseable latex formula:

\large \tex \frac{dv}{dx} = x

Unparseable latex formula:

\large \tex v = \frac{x^2}{2}

Unparseable latex formula:

\large \tex \int xlnx = \frac{1}{2}x^2 lnx - \int \frac{x^2}{2x}

Unparseable latex formula:

\large \tex \int xlnx = \frac{1}{2}x^2lnx - \int \frac{x}{2}

Unparseable latex formula:

\large \tex \int xlnx = \frac{1}{2}x^2lnx - \frac{1}{4}x^2 + C

y = xlnx

\frac{dy}{dx} = (x)\frac{1}{x} + lnx

\frac{dy}{dx} = 1 + lnx

1. Differentiate x^x

Unparseable latex formula:

\large \tex y = x^x

Unparseable latex formula:

\large \tex ln y = xlnx

Unparseable latex formula:

\large \tex \frac{1}{y}\frac{dy}{dx} = 1 + lnx

Unparseable latex formula:

\large \tex \frac{dy}{dx} = (1+lnx)y

2. Integrate cos^2(x) and cos^3(x).

Unparseable latex formula:

\large \tex \int cos^2x

From the double angle formula

Unparseable latex formula:

\large \tex 2cos^2x -1 = cos2x

Unparseable latex formula:

\large \tex cos^2x = \frac{1}{2}(1 + cos2x)

Unparseable latex formula:

\large \tex \int \frac{1}{2} + \frac{1}{2}cos2x

Unparseable latex formula:

\large \tex = \frac{1}{2}x + \frac{1}{4}sin2x + c

Unparseable latex formula:

\large \tex \int cos^3x

Unparseable latex formula:

\large \tex \int (cos^2x)cosx

Unparseable latex formula:

\large \tex \int (1-sin^2x)cosx

Unparseable latex formula:

\large \tex \int cosx - sin^2xcosx

Unparseable latex formula:

\large \tex sinx - \frac{1}{3}sin^3x + c

Using the formula

Unparseable latex formula:

\large \tex \int sin^nxcosx = \frac{1}{n+1}sin^{n+1}x + c

Felt like doing some calculus tonight.

Reply 29

dvs

Lusus Naturae

Q87. Show n^5 - n^3 is divisible by 12.

Hmm, messy way, surely there must be better:
[...]

You can tidy up your solution as follows:

n^5 - n^3 = n^3 (n-1)(n+1)

If n is odd, then n-1 and n+1 are even, and so their product is divisible by 4. If n is even, then n^2 is divisible by 4.

Now one of n-1, n+1 and n must be divisible by 3. So in both cases the n^5 - n^3 is divisible by 3*4 = 12.

Reply 30

Oldmannow

wow..those are quite hard.

i doubt many in my class would get these (i can see quite a few that we could but most of them are :redface:

)

Reply 31

dvs

insparato

3. What is the square root of i?

I guess

Unparseable latex formula:

\large \tex i = \sqrt{-1}

Unparseable latex formula:

\large \tex \sqrt i = (\sqrt{-1}^{\frac{1}{2}} = (-1)^{\frac{1}{4}}

Well, yes, but what's that equal to?

HINT: i = e^(i(pi/2) + 2kpi) and use Euler's formula.

Reply 32

zrancis

insparato

3. What is the square root of i?

I guess

Unparseable latex formula:

\large \tex i = \sqrt{-1}

Unparseable latex formula:

\large \tex \sqrt i = (\sqrt{-1}^{\frac{1}{2}} = (-1)^{\frac{1}{4}}

Argh no you can't do that! It looks all horrible!

Root i should look nice, it's the rule.

Take a look at El Matematico's solution earlier on in the post

Reply 33

insparato

ahhhh you see havent done that yet :p:

. I was just having a stab lol.

Suppose im lucky im not applying to do maths at uni :biggrin:

Reply 34

zrancis

insparato

ahhhh you see havent done that yet :p:

. I was just having a stab lol.

Another way to try it is to use the method:

(a + ib)^2 = i

Compare real and imaginary coefficents and solve simultaneously. It doesn't come out as neat as in De Moivres form but it's the right answer all the same!

Reply 35

dvs

zrancis, you should keep in mind there are infinitely many solutions for z^2 = i.

Reply 36

zrancis

dvs

zrancis, you should keep in mind there are infinitely many solutions for z^2 = i.

Really? Could you explain to me how that works? :confused:

I am aware that the method I mentioned does work so I'm not too sure how that affects the answer?

Reply 37

Goldenratio

They asked me quite a bit on combinatronics, ie map colouring and graph colouring etc.... Look up your professor's field of interest, the questions might lean towards it. If the professor gives you a list and asks you to choose ones to do... go do the ones you CAN do (obviously)!

Reply 38

insparato

zrancis

Me thinks De moivres theorem comes up in FP3 for moi :smile:

. Ive briefly looked at it, it kinda mentions it briefly in FP1 but leaves it hanging.

Reply 39

zrancis

insparato

Me thinks De moivres theorem comes up in FP3 for moi :smile:

. Ive briefly looked at it, it kinda mentions it briefly in FP1 but leaves it hanging.

Yeah, basically all it is is another way of writing complex numbers.

e^(itheta) = cos(theta) + isin(theta)

root i = e^i(pi/4)

= cos(pi/4) + isin(pi/4)

=(sqrt2)/2*(1+i)