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    I got e for number 10 MC--I believe the answer was if PI(n) = 2, you can get every last digit, which I thought was false because you can't get 4, because 2 is the only even prime. I believe the question was asking which answer choice was false.
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    (Original post by pressurize)
    I got e for number 10 MC--I believe the answer was if PI(n) = 2, you can get every last digit, which I thought was false because you can't get 4, because 2 is the only even prime. I believe the question was asking which answer choice was false.
    isn't P(14) = 2 (since 14 = 2 * 7)
    and x(14) = 4

    ?

    or did i misread what P(x) was
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    (Original post by pressurize)
    I got e for number 10 MC--I believe the answer was if PI(n) = 2, you can get every last digit, which I thought was false because you can't get 4, because 2 is the only even prime. I believe the question was asking which answer choice was false.
    if you try 11*4 you get 2 distinct factors
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    You're right...oops
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    (Original post by m0.4444)
    if you try 11*4 you get 2 distinct factors
    The answer to j was b
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    (Original post by Amitbalter)
    The answer to j was b
    I think i put b, not 100% sure.
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    (Original post by pressurize)
    You're right...oops
    Ive made tons of silly mistakes too. There more annoying when you understand the question but mess it up anyways.
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    Is it possible to get partial credit on a specific part in a long problem? slash is 80 a decent score (im american)
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    (Original post by Amitbalter)
    The answer to j was b
    Yes i got b
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    (Original post by pressurize)
    Is it possible to get partial credit on a specific part in a long problem? slash is 80 a decent score (im american)
    80 will be a fantastic score, and yes.
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    Well, after looking at this I think I got around 60, hopefully it's good enough
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    (Original post by Quido)
    Well, after looking at this I think I got around 60, hopefully it's good enough
    I think I got around 65 to 70. I hope it doesn't go down any further as this forum gets longer.
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    (Original post by m0.4444)
    I think I got around 65 to 70. I hope it doesn't go down any further as this forum gets longer.
    I'm just gonna have to avoid this thread for a while haha
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    Is anyone gonna upload an unofficial mark scheme?
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    (Original post by Quido)
    I'm just gonna have to avoid this thread for a while haha
    I am pretty much non-productive unless i see this thread. But I wish I would just stay away from it.
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    (Original post by RuairiMorrissey)
    I got 10 combinations but 107 and 117 for c? I couldn't explain why they were the values though.
    If you calculate them all carefully, they all come to 107. This rhymes with the fact that the function A^n(B^m(x))= (2^n)(3^n)x+(2^n*3^n - 1) which is used in the last question. Note that 107 = 108-1

    The latter also shows that it is impossible to reach 214x+96 (or whatever it was) since for each compound function added, the constant is > 1/2 the x coefficient. Hence, since 96 < 1/2 of 214, it is impossible with any sum.
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    (Original post by DavidBick)
    If you calculate them all carefully, they all come to 107. This rhymes with the fact that the function A^n(B^m(x))= (2^n)(3^n)x+(2^n*3^n - 1) which is used in the last question. Note that 107 = 108-1

    The latter also shows that it is impossible to reach 214x+96 (or whatever it was) since for each compound function added, the constant is > 1/2 the x coefficient. Hence, since 96 < 1/2 of 214, it is impossible with any sum.
    How many marks do you think those two parts were?
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    (Original post by DavidBick)
    If you calculate them all carefully, they all come to 107. This rhymes with the fact that the function A^n(B^m(x))= (2^n)(3^n)x+(2^n*3^n - 1) which is used in the last question. Note that 107 = 108-1

    The latter also shows that it is impossible to reach 214x+96 (or whatever it was) since for each compound function added, the constant is > 1/2 the x coefficient. Hence, since 96 < 1/2 of 214, it is impossible with any sum.
    They didn't all come to 107.
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    (Original post by DavidBick)
    If you calculate them all carefully, they all come to 107. This rhymes with the fact that the function A^n(B^m(x))= (2^n)(3^n)x+(2^n*3^n - 1) which is used in the last question. Note that 107 = 108-1

    The latter also shows that it is impossible to reach 214x+96 (or whatever it was) since for each compound function added, the constant is > 1/2 the x coefficient. Hence, since 96 < 1/2 of 214, it is impossible with any sum.
    I got the same result but said 1/6 instead of 1/2 since m and n both are positive ints.
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    I feel like I've not done well enough to get an interview : (
 
 
 
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