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    Hi there,

    I'm really stuck with these two questions despite having been given a few hints. It's question 4 and 5 of this sheet: https://www.dpmms.cam.ac.uk/study/IA...umset12012.pdf

    For question 4 I've been told that you can generate a set of non-primes using two consecutive primes e.g 5 and 7. Then 2*3*5=30.....Sequence 32,33,34,35,36 will be a sequence of (7-2) = 5 non primes. However, I tried this with 11 and 13 and it doesn't seem to work (2*3*5*7*11 = 2310, numbers from 2312 to 2332 are non-primes which isn't 13-2 = 11 numbers long). Could anyone explain this theory in more detail please?

    For question 5, I've been told to take mod 3, however, I'm confused at how you take the modulo of a sum of two numbers. Plus, I don't know how you could know the modulo of these two numbers without working it out on a calculator (except mod 5 and mod 2). Any help would be much appreciated.

    Thanks,
    Dan
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    What you've been told for question 4 is plain and simply incorrect. (More likely that you've got confused and they didn't actually say that, to be honest).

    However, given N, suppose k divides N. Then 2 also divides N+k, so N+k isn't prime.

    Now, if you know N+2, N+3, N+4, N+101 are all non-prime, you're done. So you want to choose N such that there are a *lot* of possibilities for k dividing N...

    As far as adding, subtracting or multiplying modulo M (not necessarily prime), the basic rules are:

    (a + b) mod M = [ (a mod M) + (b mod M) ] mod M.

    ab mod M = (a mod M)(b mod M) mod M.

    (In this context, "a mod M" means the remainder after dividing a by M).
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    (Original post by DanKeitley)
    ...
    Q4: factorials. It's hard to give a supplementary hint without giving it away. (n+1)! has at the very least, n factors (excluding 1); find a sequence where each term is divisible by one of these factors.

    Q5: work with each term separately. 2^{3}\equiv -1\pmod{3}\Rightarrow 2^{18}\equiv 1\pmod{3}\Rightarrow \cdots

    Do the same for 5^{40}

    Edit: Ach! Too slow
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    (Original post by DanKeitley)
    However, I tried this with 11 and 13 and it doesn't seem to work (2*3*5*7*11 = 2310, numbers from 2312 to 2332 are non-primes which isn't 13-2 = 11 numbers long).
    You mean 2312 to 2322 and it is 11 numbers long.

    - the easiest bit!
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    (Original post by DFranklin)
    What you've been told for question 4 is plain and simply incorrect. (More likely that you've got confused and they didn't actually say that, to be honest).

    However, given N, suppose k divides N. Then 2 also divides N+k, so N+k isn't prime.

    Now, if you know N+2, N+3, N+4, N+101 are all non-prime, you're done. So you want to choose N such that there are a *lot* of possibilities for k dividing N...
    Choosing the value of N is the difficult part! The hint I received was from my Maths teacher who has studied in Cambridge and I copied what he said in an email directly onto here. Are you sure it is incorrect? It works with 7 and 11 as well.

    (Original post by Lord of the Flies)
    ....
    How did you go from knowing the mod for 2^3 to knowing the mod of 2^18? I haven't been taught any of this so I apologise if you're spelling this out for me.

    (Original post by ghostwalker)
    You mean 2312 to 2322 and it is 11 numbers long.

    - the easiest bit!
    I don't understand. 2323 is divisible by 101. The next prime number is 2332. Unless it means that the gap is at least 11 numbers long...
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    I don't understand. 2323 is divisible by 101. The next prime number is 2332. Unless it means that the gap is at least 11 numbers long...
    Yes, at least....
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    (Original post by DanKeitley)
    Choosing the value of N is the difficult part! The hint I received was from my Maths teacher who has studied in Cambridge and I copied what he said in an email directly onto here. Are you sure it is incorrect? It works with 7 and 11 as well.



    How did you go from knowing the mod for 2^3 to knowing the mod of 2^18? I haven't been taught any of this so I apologise if you're spelling this out for me.



    I don't understand. 2323 is divisible by 101. The next prime number is 2332. Unless it means that the gap is at least 11 numbers long...
    Because if you have 2^{3}\equiv -1\pmod{3} then 2^{3n} \equiv (-1)^n \pmod{3}
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    (Original post by Noble.)
    Because if you have 2^{3}\equiv -1\pmod{3} then 2^{3n} \equiv (-1)^n \pmod{3}
    Ah that clears a lot of things up! Thanks
 
 
 
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