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# Oxford MAT 2013/2014 Watch

1. (Original post by nZeac)
I got the same thing for iii), but I'm not sure about ii. If she has 3 and Bob has 2, she knows she has 3 after looking at Bob because it can't be 1. I thought it was Alice: 5

In both cases (Bob: 4/Bob:6), she would have 2 primes to choose from, therefore she doesn't know.
Yep and Alice, being the great logician that she is, would know that Bob therefore must be 4 if she was 3 by deducing from your reason above.
2. (Original post by pranay1995)
(Original post by Yezi_L)
I just got:
ii) Alice:3 and Bob:2
iii) Alice is 4

Correct me if I'm wrong
(Original post by yl95)
I thought Alice was 5?

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(Original post by nZeac)
I got the same thing for iii), but I'm not sure about ii. If she has 3 and Bob has 2, she knows she has 3 after looking at Bob because it can't be 1. I thought it was Alice: 5

In both cases (Bob: 4/Bob:6), she would have 2 primes to choose from, therefore she doesn't know.
Math applicant, but I always like Q6! Here's what I got:
Spoiler:
Show
i. If B is not equal to 1, then A can be both B+1 as well as B-1. As Alice is being able to uniquely determine A, then B=1 and A=2.

ii. B must be between two prime numbers, otherwise Alice would know A. Therefore, B=4/6. Then A=3/5/7. If A=3/7, Bob would know that B=4/6. As he can't say anything, A=5.

iii. B is not 1 or 10. We can also deduce that B+1 and B-1 aren't both square/non-square. One must be a square number, the other must be a non square number. So B can be 3, 5 or 8. Then A can be 2/4, 4/6, 7/9. As Bob doesn't know what B is even after seeing A, A=4.
3. (Original post by souktik)
Math applicant, but I always like Q6! Here's what I got:
Spoiler:
Show
i. If B is not equal to 1, then A can be both B+1 as well as B-1. As Alice is being able to uniquely determine A, then B=1 and A=2.

ii. B must be between two prime numbers, otherwise Alice would know A. Therefore, B=4/6. Then A=3/5/7. If A=3/7, Bob would know that B=4/6. As he can't say anything, A=5.

iii. B is not 1 or 10. We can also deduce that B+1 and B-1 aren't both square/non-square. One must be a square number, the other must be a non square number. So B can be 3, 5 or 8. Then A can be 2/4, 4/6, 7/9. As Bob doesn't know what B is even after seeing A, A=4.
can you confirm answers to q5?
4. (Original post by pranay1995)
can you confirm answers to q5?
9, n+1, (n+1)(n+2)/2, 10, 30, 13500.
5. (Original post by yl95)
I thought Alice was 5?

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(Original post by nZeac)
I got the same thing for iii), but I'm not sure about ii. If she has 3 and Bob has 2, she knows she has 3 after looking at Bob because it can't be 1. I thought it was Alice: 5

In both cases (Bob: 4/Bob:6), she would have 2 primes to choose from, therefore she doesn't know.
(Original post by souktik)
Math applicant, but I always like Q6! Here's what I got:
Spoiler:
Show
i. If B is not equal to 1, then A can be both B+1 as well as B-1. As Alice is being able to uniquely determine A, then B=1 and A=2.

ii. B must be between two prime numbers, otherwise Alice would know A. Therefore, B=4/6. Then A=3/5/7. If A=3/7, Bob would know that B=4/6. As he can't say anything, A=5.

iii. B is not 1 or 10. We can also deduce that B+1 and B-1 aren't both square/non-square. One must be a square number, the other must be a non square number. So B can be 3, 5 or 8. Then A can be 2/4, 4/6, 7/9. As Bob doesn't know what B is even after seeing A, A=4.
Oops! Sorry for that silly mistake, Alice is 5 for ii). You're all right

Sadly, I didn't get to do this question either...
6. Is the cutoff for only computer science applicants always high?
7. (Original post by souktik)
9, n+1, (n+1)(n+2)/2, 10, 30, 13500.
Soutik I got all of those except 30- it really is unfair because that question was ambiguous. I thought it meant there was at least one 5 in it. so..

5_ _
_ 5 _
_ _ 5

are only configurations. 4 ways to add up to three each since total is 8

so 4*3=12

If I lose marks for this it is really unfair.
8. For the last multiple choice, how come it wasn't just a geometric series? If you sketch out the graph of floor(2^x) it's a staircase starting at 1 and increasing exponentially (i.e. it takes values of 1,2,4,8), and furthermore, all of those steps last for one unit on the x-axis. So, interpreting the integral as the area under the curve, you add up all those bars, and would it not just be the sum from 0 to n of 2^n, which equals (2^n)-1?
9. (Original post by souktik)
9, n+1, (n+1)(n+2)/2, 10, 30, 13500.
What do you think the mark scheme for this question would be? Would it be fair to say that I can get 7 points for parts 1,2 and 6?
10. (Original post by nahomyemane778)
Soutik I got all of those except 30- it really is unfair because that question was ambiguous. I thought it meant there was at least one 5 in it. so..

5_ _
_ 5 _
_ _ 5

are only configurations. 4 ways to add up to three each since total is 8

so 4*3=12

If I lose marks for this it is really unfair.
I sympathize, but I don't think they'll award you marks. It probably carries just a couple of marks, so I don't think it's going to make a huge difference either.

(Original post by 'murica1776)
What do you think the mark scheme for this question would be? Would it be fair to say that I can get 7 points for parts 1,2 and 6?
Yeah, you'll probably get 7 or 8. As for your doubt regarding the MCQ, I would like to point out that the function can take values between two powers of 2 as well. For example, the value is 3 between x=log_2(3) and log_2(4).

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11. (Original post by 'murica1776)
For the last multiple choice, how come it wasn't just a geometric series? If you sketch out the graph of floor(2^x) it's a staircase starting at 1 and increasing exponentially (i.e. it takes values of 1,2,4,8), and furthermore, all of those steps last for one unit on the x-axis. So, interpreting the integral as the area under the curve, you add up all those bars, and would it not just be the sum from 0 to n of 2^n, which equals (2^n)-1?
Because the floor function operates on the output, therefore the y-axis needs to go up in integers therefore the x-axis involves logs.
12. (Original post by Noble.)
...
Noble- when will we be invited to interview?
13. (Original post by nahomyemane778)
Noble- when will we be invited to interview?
End of November and early December. Expect to hear back ~2 weeks before interviews begin.
14. (Original post by Tarquin Digby)
https://www.maths.ox.ac.uk/system/fi...nts/test13.pdf

Paper's up. I got ACCBDADBBB for multiple choice.
Seems this post has been now been edited to give all the right answers. Does this mean you have a hot line to the markers to change your answers as well?
15. any got hold of the paper yet if so please link me to it
16. (Original post by TheGoldenRatio)
any got hold of the paper yet if so please link me to it
Look at the post just above yours.

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17. (Original post by souktik)
Look at the post just above yours.

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fml, i am soooo blind! , though any unofficial solutions in one please rather than all over the place!
18. (Original post by TheGoldenRatio)
fml, i am soooo blind! , though any unofficial solutions in one please rather than all over the place!
Sorry, I don't think you'll get that.
If you go through my posts you'll find solutions to problems 2, 4 and 5. I don't think I posted 3, but some others did. In case you applied to one of the CS courses, I even posted the solution to problem 6.

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19. (Original post by Noble.)
End of November and early December. Expect to hear back ~2 weeks before interviews begin.
Merci.

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