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# Trinity Admissions Test Solutions watch

1. Jordan\ Youshould also add the solutions for the downing college one for more practice

http://www.dow.cam.ac.uk/documents/t...Maths_test.pdf
2. (Original post by FutureOxford3)
Jordan\ Youshould also add the solutions for the downing college one for more practice

http://www.dow.cam.ac.uk/documents/t...Maths_test.pdf
Yeah, of course! I'll add it to the OP.
3. (Original post by Zacken)
Very much so, the questions test general maths and will be useful practice for any college including Oxford.
Oxford is not a college tho :P
4. (Original post by Krollo)
Two pleasingly different solutions. What maths is all about, perhaps

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The final answer is correct but after spotting that pattern it has been assumed it holds for all n, I would prove it by induction then let n=1000.

I can't found who did the A one, but I can't find so maybe Krollo can make me find the solution-er.
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5. (Original post by physicsmaths)
The final answer is correct but after spotting that pattern it has been assumed it holds for all n, I would prove it by induction then let n=1000.

I can't found who did the A one, but I can't find so maybe Krollo can make me find the solution-er.
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Hence why I deleted it
6. (Original post by Jordan\)
Hence why I deleted it
I thought i just could not find it. Assume it holds for. K, show it holds for k+1 (by multiplying by A) and that's it.

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7. Did that Z(root2) question on paper, I will write it up soon whilst I learn latex..

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8. (Original post by physicsmaths)
Did that Z(root2) question on paper, I will write it up soon whilst I learn latex..

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Pell equation!

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9. (Original post by Renzhi10122)
Pell equation!

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Yep! Amazing stuff that is

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10. (Original post by FutureOxford3)
Jordan\ Youshould also add the solutions for the downing college one for more practice

http://www.dow.cam.ac.uk/documents/t...Maths_test.pdf
We figured those were far too elementary to provide solutions for. You can check your answers with a quick WA computation.
11. (Original post by Zacken)
We figured those were far too elementary to provide solutions for. You can check yoir answers with a quick WA computation.
What is the answer to 5 and 6 then?
12. (Original post by FutureOxford3)
What is the answer to 5 and 6 then?
Don't have the paper with me, but the question is a straightforward sketch and calculus problem. You can check your answers using WA.

Six is admittedly hard, but it's not worth providing 8 solutions if we only really need it for 1 question.
13. (Original post by Zacken)
Don't have the paper with me, but the question is a straightforward sketch and calculus problem. You can check your answers using WA.

Six is admittedly hard, but it's not worth providing 8 solutions if we only really need it for 1 question.
Yea, the only one I can't do is number 6. I will just ask Jordan or physicsmath to help me then.
14. (Original post by Krollo)
Specimen 2, Q7

Hint
Spoiler:
Show

(x-1) is a factor of x^n -1 always.
When is (x+1) a factor of x^n+1?

Solution
Spoiler:
Show

Unless otherwise stated, assume all new constants are positive integers.

Suppose p = a^n - 1.
Then p = (a-1)(a^n-1 + ... + 1)

So p is composite unless a-1 = 1 (as the second bracket is greater or equal to the first), so a=2.

So p = 2^n - 1.

Suppose n is composite, i.e. n=cd, with neither c or d less than 2.

Then p =l 2^(cd) -1 = (2^c)^d -1 = (2^c -1)(2^(cd-c) + ... +1)

As c>1, p is thus composite. So n cannot be composite - in other words, n is prime.

- - -

Let q = a^n + 1. We seek the conditions in which q is prime.

If a is 1, then q is clearly always prime.

If a is an odd integer greater than 1, then a^n is odd, so q is even (and not 2) hence it is never prime in this case.

If a is an even integer, then if n is odd and greater than 1, q=(a+1)(a^n-1 - ... +1). q is thus never prime as a+1 > 1.

If a is an even integer, then if n is even but has an odd factor r where r>1, and n=rs, then q = a^(rs) +1 = (a^s)^r + 1 = (a^s+1)(a^rs-s... +1) and so q is never prime.

This leaves the only possibility as when a is even, and n is even and has no odd factors > 1, i.e. n is a non-negative integer power of two. Examples of this would be
2^4 +1 =17, 6^1+1 = 7, 6^2 +1 =37.

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Um, didn't the question ask for 2^n+1 not a^n+1? Also, when a = 8 n =2 you get 65.
15. (Original post by ForeignStudent34)
Um, didn't the question ask for 2^n+1 not a^n+1? Also, when a = 8 n =2 you get 65.
Well, they tell you to look at a^n -1 first and see what you can notice, they then want you to notice a form of a^n+1
And then factor and see what you can see when a=2. I have not read through Krollos solution but i trust it is correct as his structure/idea looks fine imo.

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16. I think in general they want to see 2^n+1=2^n-(-1)=(3)(some integer) if n is odd 2^n+1 is prime only when n=1 then do a similar thing for n is even, which krollo has done but to a greater depth.

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17. Specimen Test 2, Question 5:

Hint:
Spoiler:
Show
Use the equation
Solution:
Spoiler:
Show
a) Let be the event the final coin selected is biased and let be the event the coin comes up heads three times

b) Let be the event the biased coin is among the three selected and let be the event the coin comes up heads three times

Question 2:
i)
Spoiler:
Show

ii)
Spoiler:
Show

or

Question 3:
Hint:
Spoiler:
Show
Think about how taking -1 as a factor in a modulus function affects the function
Solution:
Spoiler:
Show
Because both inequalities are less than E/2 we may safely add them:

(1)

(2)
Question 4:
Spoiler:
Show
Let V denote the volume:

or

x cannot be so x must be

Check:

When x = :

Therefore V is a maximum at x =

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