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Six incredibly difficult maths puzzles that only the world’s elite can answer

Man solves puzzle from the International Maths Olympiad

Man solves puzzle from the International Maths Olympiad

How good are you at math (or ‘maths’ if you live in the UK)? If you’re at all mathematically accomplished, you’ll easily answer the international PISA Test questions aimed at 15 year olds globally. You’ll also easily be able to answer one of the more difficult questions from this year’s GCSE exams aimed at 16 year-olds in the UK.  But can you answer the kinds of maths puzzles encountered by the world’s best mathematicians in the International Mathematical Olympiad which started yesterday?

We’ve provided a selection of puzzles from Olympiad papers over the past nine years on the page below. If you think you have the solutions, please submit them via the comments box at the bottom of this page. We’re not able to offer a prize, but we will be suitably impressed.

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Olympiad 1

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Comments (9)

Comments
  1. Q5.

    If a and b are both positive integers, and (4a^(2)-1)^(2) is divisible by 4ab – 1, we can say that (4a^(2)-1)^(2) / 4ab – 1 = k, for some even positive integer k.

    Expanding the numerator, we get (4a^(2)-1)(4a^(2)-1) / 4ab – 1 = k

    i.e. (4a^(2)-1)(4a^(2)-1) = k(4ab – 1)

    So we can deduce that (4a^(2)-1) = k, (4a^(2)-1) = 4ab – 1.

    Taking the second instance, (4a^(2)-1) = 4ab – 1, we can reduce this down to a = b, where a and b are positive integers.

    Putting this into the remaining equation, (4a^(2)-1) = k, we get 4•a•a – 1 = k, so when a = b, we get 4ab – 1 = k, which satisfies the original equation.

  2. it is not because X^2 = YZ that X=Y and X=Z… take 6^2 = 4*9

  3. It is not because X^2 = YZ that X=Y and X=Z… take 6^2 = 36 = 4*9

  4. I have discovered a truly marvelous proof of all these questions, which this margin is too narrow to contain.

  5. You’re right, that was lazy. Try this?

    (4a^(2) – 1)^2 / 4ab – 1 = k

    (4a^(2) – 1)^2 = k(4ab – 1)

    16a^4 – 8a^2 + 1 = 4abk – k

    16a^4 – 8a^2 – 4abk + k + 1 = 0

    4a(a^3 – 2a – bk) + k + 1 = 0

    4a(a^3 – 2a – bk) = -(k+1)

    i.e. 4a = -(k+1), a^3 – 2a – bk = -(k+1)

    Taking 4a = -(k+1), and letting (k+1) = A:

    (-A^(2) – 1)^2 = (A-1)(4ab -1)

    A^4 – 2A^(2) + 1 = (A-1)(4ab – 1)

    A^4 – 2A^(2) + 1 / (A-1) = 4ab – 1

    Divide through with long division and you’ll get A^3 – A^2 – A + 1 = 4ab – 1

    i.e. A^3 – A^2 – A = 4ab

    i.e. 4ab = A(A^2 – A – 1)

    so 4ab = k+1, (k+1)^2 – k

    As k is the result of division, k>0.

    Let k = 1 for the sake of argument in the latter case: 4ab = 0 => a = b = 0, which is impossible given our constraint

    In the former case: k = 1 => 4ab = 2, which works.

    Hence we must focus on the case where 4ab = k + 1.

    Now let 4ab = k + 1, where k>0

    Recalling our equation such that (4a^(2) – 1)^2 / (4ab – 1) = k, now:

    (4a^(2) – 1)^2 = k(k+1-1) = k^2

    so (4a^(2) – 1)^2 = k^2

    Square root both sides: 4a^2 – 1 = k

    4a^2 = k+1

    a^2 = k+1 / 4

    a = + √(k+1)/2 (we take the positive root as a>0)

    Using 4ab = k+1:

    4 • [√(k+1)/2] • b = k+1

    2 • √(k+1) • b = k+1

    2b • √(k+1) = k+1

    2b = (k+1) / (k+1)^(1/2) = (k+1)^(1/2)

    Thus, b = √(k+1)/2, and a = b, where a, b > 0, for some k>0

  6. to MakingAPunt there flawed points in your solution not sure they work
    Q5:
    Let’s assume p prime number and it be p^h the highest power of p that divides 4ab−1 (a
    n odd number). Hence, we infer that p^h is odd and it divides e (4a^2 − 1)^2 = (2a − 1)^2
    (2a + 1)^2. Given that (2a + 1)-(2a − 1)=2 then their greatest common divisor is 2. Therefore, p^h can only divide either (2a − 1)^2 or (2a + 1)^2. It be 2k=h if h is even or 2k=h if h is odd then p^2k can only divide either (2a − 1)^2 or (2a + 1)^2 so p^k can only divide either (2a − 1) or (2a + 1). We have now that 2a ≡ ±1 mod p^k . If a ≡ 1 mod p^k in that case we have 0 ≡ 4ab − 1 =
    2a · 2b − 1 ≡ 2b − 1 mod p^k from which we can infer that 2b ≡ 1 ≡ a mod p^k. If instead a ≡ −1 mod p^k in that case 0 ≡ 4ab − 1 = 2a · 2b − 1 ≡ −2b − 1 mod p^k from which we can infer that e 2b ≡ −1 ≡ a mod p^k. In any case we found that a ≡ b mod p^k.

    QED

  7. @makingapunt
    you said that 16a^4 – 8a^2 – 4abk + k + 1 = 0 and then simplify it to 4a(a^3 – 2a – bk) + k + 1 = 0, should not the simplification be 4a(4a^3 – 2a – bk) + k + 1=0.
    and you said that A^4 – 2A^(2) + 1 / (A-1) = 4ab – 1 is just A^3 – A^2 – A + 1 = 4ab – 1, by multiplying the equation with a-1 you won’t be back at the innate equation. otherwise i did enjoy your workout really ingenious.

  8. problem 2
    we a2.a3…..an=1, we can rewrite this as an ! / a1! = an.an-1.an-2……..a3.a2.a1 / a1, this is saying an factorial divide it by a1 factorial equal , as we know that a is a positive real integers thus 1 is the smallest result that a factorial could get, which make infer that a2.a3…..an=1 is only possible if and only if an=an-1=an-2=….=a2=1.
    lets prove that (1+a2)^2(1+a3)^3….(1+an)^n > n^n , as we know that a2.a3…..an=1 then (1+a2)^2(1+a3)^3….(1+an)^n = (1+a)^n+n-1+….+2 = (1+a)^(((n^2+n)/2)-1).
    so the equation becomes (1+a)^(((n^2+n)/2)-1) > (n^n), as a=1, we have
    2^((n^2+n)/2)>2(n^n),by dividing both side by 1/n we have 2^((n+1)/2)>2n,hence for all n > 3 this equation is satisfied

  9. problem 2 – ibradiouf

    so far we only know that ai are real numbers (not necessarilly integrers), also it’s trivial to find positive real numbers ai so that their product equals 1 (if you take rationals for instance)

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