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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.
    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.
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    (Original post by pranay1995)
    Computer science applicants..what were your answers to q6..
    (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.
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    (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?
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    (Original post by pranay1995)
    can you confirm answers to q5?
    9, n+1, (n+1)(n+2)/2, 10, 30, 13500.
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    Maths


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    (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...
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    Is the cutoff for only computer science applicants always high?
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    (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.
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    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?
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    (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?
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    (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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    (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.
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    (Original post by Noble.)
    ...
    Noble- when will we be invited to interview?
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    (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.
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    (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?
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    any got hold of the paper yet if so please link me to it
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    (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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    (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!
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    (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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    (Original post by Noble.)
    End of November and early December. Expect to hear back ~2 weeks before interviews begin.
    Merci.
 
 
 
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