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    (Original post by Jkn)
    Classic science teachers :')

    Well I was only every really interested in theoretical physics and 'the big questions' (sheldon-cooper style, standard) and I met some current student at my open day, one from maths and one from natsci, and they said that maths was 100% the best route into it and that even 'maths with physics' in my first year would hold me back!

    That was the first step and since then I've found that the beauty and satisfaction in pure maths is immense and is completely different to everything else I've experienced. Besides, I'm not really fussed about money and I don't want to dedicate my life to coming up with things that makes peoples lives easier, so why not explore the cosmos/ mathematical universe!
    Why did they say that?

    I can understand why that is attractive. I'd enjoy the practical experiments as well as the theoretical side. Whilst the money isn't my ultimate desire, the fact that I'd enjoy both Chem Eng and Physics means that it does play a part in my decision.
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    (Original post by Jkn)
    Just remembered a really nice STEP question I did while ago and thought of this...

    Problem 88*

    P(N) is defined as the largest product that can be made from positive integers that add up to N.

    i.e. P(N)=max(a_{1}a_{2}...a_{n}) : N= a_{1}+a_{2}+...+a_{n}.

    Prove the Goldbach Conjecture
    Spoiler:
    Show
    lol jk find P(N) :lol:

    Solution 88*

    By the AM-GM inequality we have

    (a_{1}a_{2}...a_{n})^{1/n} \leq \frac{a_{1}+a_{2}+...+a_{n}}{n}

    so

    a_{1}a_{2}...a_{n}=<(\frac{a_{1}  +a_{2}+...+a_{n}}{n})^n

    we have equality iff ai=aj for all i,j

    hence max(a_{1}a_{2}...a_{n})=(\frac{N  }{n})^n

    EDIT: just realised I need to maximise this as n is not fixed.
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    Solution 89

    Both A andB are prime numbers. Thus, B=7 \times 6+1=43, and A=173. Hence N=347.
    Spoiler:
    Show
    The case B=29 leads to A\ge 59, and N \ge 467


    Solution 88

    Obviously, we have to partition into parts which are < 5. Notice that 2^{3}<3^{2}, thus we can assume that we have at most two 2's and the remaining number are 3's (2^{2}=2+2).
    If N =3m, P(N)=3^{m}.
    If N = 3m+1, P(N)=3^{m-1}.2^{2}
    If N =3m+2, P(N)=3^{m}2.
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    (Original post by james22)
    max(a_{1}a_{2}...a_{n})=(\frac{N  }{n})^n

    EDIT: just realised I need to maximise this as n is not fixed.
    I just thought of an awesome way to adapt this
    Spoiler:
    Show
    Try writing N/n=x and use some calculus


    (Original post by Mladenov)
    Solution 89

    Both A andB are prime numbers. Thus, B=7 \times 6+1=43, and A=173. Hence N=347.
    Spoiler:
    Show
    The case B=29 leads to A\ge 59, and N \ge 467
    Correct Am I right in saying you have obtained these numbers through systematic exhaustion? It would be great to find an elegant solution (I wonder if it exists...)
    Solution 88

    Obviously, we have to partition into parts which are \le 5. Notice that 2^{3}<3^{2}, thus we can assume that we have at most two 2's and the remaining number are 3's (2^{2}=2+2).
    If N =3m, P(N)=3^{m}.
    If N = 3m+1, P(N)=3^{m-1}.2^{2}
    If N =3m+2, P(N)=3^{m}2.
    Correcto, but why is the first part 'obvious'?
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    [QUOTE=Jkn;42397512]Fuuuuuuck I just thought of an awesome way to adapt this
    Spoiler:
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    Try writing N/n=x and use some calculus


    I think my entire solution is rubbish, I just re-read the question and noticed that the a_i have to be integers so my answer does not work.
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    (Original post by Jkn)
    Correct Am I right in saying you have obtained these numbers through systematic exhaustion? It would be great to find an elegant solution (I wonder if it exists...)
    You are looking for prime numbers in arithmetic progressions. I can think of some analytic number theory estimations , but then the solution is far from elementary. This problems does not seem to be elegant, since we have to deal with primes, which are greater than 100.

    (Original post by Jkn)
    Correcto, but why is the first part 'obvious'?
    Simply because 2+3=5 < 6.
    But in general, the global maximum of the curve x(a-x) occurs at \displaystyle x=\frac{a}{2}. Hence for a > 5, we can have greater product by taking b+ (a-b), where \displaystyle b= \left[\frac{a}{2}\right].
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    (Original post by james22)
    I think my entire solution is rubbish, I just re-read the question and noticed that the a_i have to be integers so my answer does not work.
    But you can have max(a_{1},a_{2}...a_{n}) \leq (\frac{N}{n})^n . The solution I have in mind eliminates the need to intuition and formalises the result by linking it to e
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    (Original post by Mladenov)
    You are looking for prime numbers in arithmetic progressions. I can think of some analytic number theory estimations , but then the solution is far from elementary. This problems does not seem to be elegant, since we have to deal with primes, which are greater than 100.
    Always a shame to use exhaustion though
    Spoiler:
    Show
    :'(

    But in general, the global maximum of the curve x(a-x) occurs at \displaystyle x=\frac{a}{2}. Hence for a \ge 5, we can have greater product by taking b+ (a-b), where \displaystyle b= [\frac{a}{2}].
    Excellent. It's the use of a fact like that that I wanted
    Spoiler:
    Show
    Also, simply considering x(a-x) \leq a is sufficient (this was that hint given in the STEP question)
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    (Original post by Jkn)
    Always a shame to use exhaustion though
    Spoiler:
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    :'(
    Absolutely. As you might have seen, I post problems with requite ideas, rather than brute force methods.

    (Original post by Jkn)
    Excellent. It's the use of a fact like that that I wanted
    Spoiler:
    Show
    Also, simply considering x(a-x) \leq a is sufficient (this was that hint given in the STEP question)
    It is quite natural to look at x(a-x), for it preserves the sum, and, in addition, you can vary the product.

    I am gonna leave problem 81 for somebody else. Yet, I have mind-blowing solution which uses abelian varieties.
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    (Original post by Mladenov)
    Absolutely. As you might have seen, I post problems with requite ideas, rather than brute force methods.

    It is quite natural to look at x(a-x), for it preserves the sum, and, in addition, you can vary the product.
    Which is what makes a good problem. Mmm, hence the possible link to AM

    Post some problems! A similar level of difficulty to 86 would be nice (or perhaps BMO2-level)
    ------------
    Wonder if anyone recognises where I got this from

    Problem 90*

    Find the sum of all values of a such that the equation  (x^2-x+a+1)^2=4a(5x^2-x+1) has 3 distinct real solutions.
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    (Original post by Jkn)
    Post some problems! A similar level of difficulty to 86 would be nice (or perhaps BMO2-level.
    Problem 91*

    Solve in integers x^{5}-x^{3}=y^{3}z, where y,z are prime numbers.

    Problem 92*

    Define a_{1}=1, and for n > 1, a_{n}=n(a_{1}+..+a_{n-1}). Then for every n \equiv 0 \pmod 2,a_{n} is divisible by n!.
    Also, find all n \equiv 1 \pmod 2 such that a_{n} \equiv 0 \pmod {n!}.
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    (Original post by Mladenov)
    Problem 91*

    Solve in integers x^{5}-x^{3}=y^{3}z, where y,z are prime numbers.
    Solution 91

    Rearranging gives x^3(x+1)(x-1)=y^3z. The RHS has 4 prime factors and so the LHS must have 4 prime factors.

    This implies one of the five terms in the product must equal 1. Noting that the expression must be positive (i.e. not equal to zero) we conclude that x=2.

    Substituting in gives y^3z=24=2^{3}.3. As y and z are prime they must match up with these number exactly. Therefore the only solution is (x,y,z)=(2,2,3)
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    (Original post by Jkn)
    Which is what makes a good problem. Mmm, hence the possible link to AM

    Post some problems! A similar level of difficulty to 86 would be nice (or perhaps BMO2-level)
    ------------
    Wonder if anyone recognises where I got this from

    Problem 90*

    Find the sum of all values of a such that the equation  (x^2-x+a+1)^2=4a(5x^2-x+1) has 3 distinct real solutions.
    Can a be complex? If not I don't think there are any a that make it have 3 distinct real solutions.
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    PS Helper
    (Original post by Jkn)
    Problem 90*
    Find the sum of all values of a such that the equation  (x^2-x+a+1)^2=4a(5x^2-x+1) has 3 distinct real solutions.
    Fairly sure this is too messy to be the right way of going about this question, but you never know:

    Take over the RHS to give (x^2-x+a+1)^2-4a(5x^2-x+1)=0
    3 distinct real solutions means that  (x^2-x+a+1)^2-4a(5x^2-x+1) = (ax^2+bx+c)(dx^2+ex+f) where b^2 > 4ac and e^2 = 4df.
    Compare coefficients to produce a sufficient number of simultaneous equations, and then solve.

    I've tried to do it on paper but got a bit lost after I found all of the equations/inequalities I could.
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    Solution 68

    Note that x\not\in\mathbb{Z}:\;[n-x]+[x]=n-1 hence letting x_i\to 1-x_i:

    \begin{aligned}\displaystyle \int_0^1 \cdots \int_{0}^1 \left[\sum_{i=1}^n x_i\right] dx_1\cdots dx_n &=\int_0^1 \cdots \int_{0}^1 \left[n-\sum_{i=1}^n x_i\right] dx_1\cdots dx_n\\&=\int_0^1 \cdots \int_{0}^1 \frac{n-1}{2}\;dx_1\cdots dx_n\\&=\frac{n-1}{2}\end{aligned}

    Solution 92

    Observe that \displaystyle 1+\big(1\cdot 1!+\cdots +n\cdot n!\big) =(n+1)! by repeatedly factoring the LHS, then it is kid's stuff to show that a_{n>1} = \frac{1}{2}n\cdot n! and obviously the only odd n for which n!|a_n is when a_n\neq \frac{1}{2}n\cdot n! so for n=1.
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    (Original post by james22)
    Can a be complex? If not I don't think there are any a that make it have 3 distinct real solutions.
    What makes you think that? :P
    (Original post by The Polymath)
    Fairly sure this is too messy to be the right way of going about this question, but you never know
    There are many ways to do problems like these, this method does seem like it could get a little messy, though seems promising
    ------------------
    (Original post by Mladenov)
    Problem 92*

    Define a_{1}=1, and for n > 1, a_{n}=n(a_{1}+..+a_{n-1}). Then for every n \equiv 0 \pmod 2,a_{n} is divisible by n!.
    Also, find all n \equiv 1 \pmod 2 such that a_{n} \equiv 0 \pmod {n!}.
    (Original post by Lord of the Flies)
    Solution 92

    Observe that \displaystyle 1+\big(1\cdot 1!+\cdots +n\cdot n!\big) =(n+1)! by repeatedly factoring the LHS, then it is kid's stuff to show that a_{n>1} = \frac{1}{2}n\cdot n! and obviously the only odd n for which n!|a_n is when a_n\neq \frac{1}{2}n\cdot n! so for n=1.
    Very nice. I took a somewhat different approach:

    Solution 92 (2)


    Clearly the statement is true for n=1 and 2. We then proceed to form an inductive argument.

    Let p=\frac{a_{n}}{n!}, where p is a positive integer.

     \frac{p(n+1)}{n}=\frac{(a_1+...+  a_{n-1})(n+1)^2}{(n+1)!}=\frac{(a_1+.  ..+a_{n-1})+n(a_1+...+a_{n-1})+n((a_1+...+a_{n-1})+n(a_1+...+a_{n-1}))}{(n+1)!}=\frac{a_{n+2}}{(n+  2)!}

    This implies (n+2)!|a_{n+2} if and only if \frac{p(n+1)}{n} is an integer. This happens in only two cases. Either n|(n+1) or n|p. The first case arises if and only if n=1 (dealing exclusively with the natural numbers, as specified). The second case implies that, if n|(n-1)! or (n-1)!|n, (i.e. n is non-prime), the inductive hypothesis is complete. On the other hand, if gcd(n, (n-1)!)=1 (i.e. n is prime), then this would imply n|(a_1+...+a_{n-1}) which cannot be true as by the definition of a_{n} and the assumption we made that n is co-prime to all proceeding positive integers.

    Hence, when considering each case base, induction is complete only complete for n=2 and not n=1 (i.e. all even numbers, with the addition of 1).

    An attempt to offer more insight into the inner working of the question (i.e. "why only even numbers plus one?") is to notice that the condition, with our particular relation(a_1=1), is only true for the base case two because it is the only such number that satisfies  (n-1)! < n .
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    (Original post by Jkn)
    Problem 88*

    P(N) is defined as the largest product that can be made from positive integers that add up to N.

    i.e. P(N)=max(a_{1}a_{2}...a_{n}) : N= a_{1}+a_{2}+...+a_{n}.

    Prove the Goldbach Conjecture
    Spoiler:
    Show
    lol jk find P(N) :lol:

    Solution 88

    Assume that a_k\geq5. Then a_k-2\geq3, and so a different combination of integers, where a_k is replaced with a_k-2 and 2, will have a greater product as 2(a_k-2)=(a_k-2)+(a_k-2)>(a_k-2)+2=a_k. Therefore a_k<5 \;\forall k

    If a_k=4, this is equivalent to a combination where a_k is replaced with 2 2s, as they have the same product and sum. So, WLOG, there must be an optimal combination with a_k<4\;\forall k. Clearly if a_k=1, a combination where another number was instead increased by 1 would have a greater product, therefore a_k\neq1. Therefore a_k=2 or 3. If there are 3 or more terms which are 2, then a combination where there are 2 3s, instead of 3 of those 2s, there will be a greater product as 2+2+2=3+3 and 2^3<3^2. Therefore the optimal combination will have only 3s and 2s, and at most 2 2s.

    Therefore, if N=0 \pmod{3}, P(N)=3^{\frac{N}{3}}, if N=1 \pmod{3}, P(N)=4\cdot3^{\frac{N-4}{3}} and if N=2 \pmod{3}, P(N)=2\cdot3^{\frac{N-2}{3}}
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    (Original post by Mladenov)
    I am gonna leave problem 81 for somebody else. Yet, I have mind-blowing solution which uses abelian varieties.
    (Only just saw you put this!)

    Sounds awesome! Message me with it if you like (though it sounds as though it is far too sophisticated for me) Btw, I haven't actually tried this problem yet, I just thought I would post it for the sake of the historical reference

    Oooh just thought of an easier version that may interest people: evaluating the case whereby the cubes do not need to be distinct
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    (Original post by The Polymath)
    Fairly sure this is too messy to be the right way of going about this question, but you never know:

    Take over the RHS to give (x^2-x+a+1)^2-4a(5x^2-x+1)=0
    3 distinct real solutions means that  (x^2-x+a+1)^2-4a(5x^2-x+1) = (ax^2+bx+c)(dx^2+ex+f) where b^2 > 4ac and e^2 = 4df.
    Compare coefficients to produce a sufficient number of simultaneous equations, and then solve.

    I've tried to do it on paper but got a bit lost after I found all of the equations/inequalities I could.
    I don't think this will work, as a quadratic must have either a single repeated root, or 2 real or complex roots, and it was explicitly stated that there must be 3 distinct real roots. I think that as you are supposed to take the sum of all values of a where 3 distinct real roots exist, there must only be a few special values of a where this is true, so you won't be able to factorise it. As to how you find these special values, I have no idea.
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    (Original post by Zephyr1011)
    I don't think this will work, as a quadratic must have either a single repeated root, or 2 real or complex roots, and it was explicitly stated that there must be 3 distinct real roots. I think that as you are supposed to take the sum of all values of a where 3 distinct real roots exist, there must only be a few special values of a where this is true, so you won't be able to factorise it. As to how you find these special values, I have no idea.
    My logic was that for there to be 3 distinct real roots of a quartic, the quartic must be equal to the product of two quadratics, one which has two distinct roots, and the other 1 repeated root, giving 3 in total.
 
 
 
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