Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Challenge Math Challenge - December 2018

  1. Nov 30, 2018 #1

    fresh_42

    User Avatar
    2017 Award

    Staff: Mentor

    It's December and we like to do a Special this month. The challenges will be posted like an Advent Calendar. We will add a new problem each day, from 12/1 to 12/25. They vary between relatively easy logical and numerical problems, calculations, to little proofs which hopefully add some nice-to-know lemmata to your toolbox. We hope you will have fun with this format. In case open questions are too difficult and a new one isn't opened yet, we recommend https://www.physicsforums.com/threads/math-challenge-november-2018.960003/#post-6087946 where still some easy ones can be found.

    Rules:

    a) In order for a solution to count, a full derivation or proof must be given. Answers with no proof will be ignored. Solutions will be posted around 15th of the following month.
    b) It is fine to use nontrivial results without proof as long as you cite them and as long as it is "common knowledge to all mathematicians". Whether the latter is satisfied will be decided on a case-by-case basis.
    c) If you have seen the problem before and remember the solution, you cannot participate in the solution to that problem.
    d) You are allowed to use google, wolframalpha or any other resource. However, you are not allowed to search the question directly. So if the question was to solve an integral, you are allowed to obtain numerical answers from software, you are allowed to search for useful integration techniques, but you cannot type in the integral in wolframalpha to see its solution.

    Hints: on demand

    1. (solved by @julian ) Let ##(a_n)_{n \in\mathbb{N}}\subseteq \mathbb{R}^+## be a sequence of positive real numbers and ##A=\sum_{n=1}^\infty a_n\,.## Prove the following statements:
    1. (Kummer) If there is a sequence ##(b_n)_{n \in \mathbb{N}}\subseteq \mathbb{R}^+## of positive real numbers, such that there is an index ##N \in\mathbb{N}## for which ##b_{n-1}\cdot \dfrac{a_{n-1}}{a_n} - b_n \geq C## for a constant ##C>0## and all ##n>N##, then ##A## converges.

    2. (Kummer) If there is a sequence ##(b_n)_{n \in \mathbb{N}}\subseteq \mathbb{R}^+## of positive real numbers, such that the series ##\sum_{n=1}^{\infty}\dfrac{1}{b_n}## diverges, and there is an index ##N\in \mathbb{N}## such that ##b_{n-1}\cdot \dfrac{a_{n-1}}{a_n}-b_n \leq 0## for all ##n>N##, then ##A## diverges.

    3. (Bertrand) We define the sequence of real numbers by $$b_n := \left(n \cdot \left( \dfrac{a_n}{a_{n+1}}-1 \right)-1 \right)\log (n)$$ and ##B :=\lim_{n \to \infty}b_n \in \overline{\mathbb{R}}=\mathbb{R}\cup \{\,\pm \infty\,\}\,.##
      Then ##A## converges if ##B>1## and diverges if ##B<1\,.##
    2. a.) (solved by @alind31 ) Here is a logic puzzle especially for our younger members:

    Two mathematicians meet by chance on a plane: "Didn't you have three sons?" asks one, "how old are they?" - "The product of years is 36," is the answer, "and the sum of years is exactly today's date." "Hmm, that's not enough for me," says the colleague. "Oh, right," says the second mathematician, "I forgot to mention that my eldest son has a dog."
    How old are the three sons?

    All others are recommended to [/URL]https://www.physicsforums.com/threads/math-challenge-november-2018.960003/. Especially the problems 7,8,9,19,20 should be manageable, and even 15,16 look more complicated as they actually are. If you want a new problem despite of these:
    2. b.) The unitary matrices build a manifold, the Lie group ##SU(n)##. Calculate its tangent space at the neutral element of the group.

    3. Find functions ##y(t)## and ##z(t)## which locally solve the equations
    $$
    \begin{cases}
    e^t + \tan y(t) &= 1\\
    t^2 + z(t)^3+z(t) &=0
    \end{cases}
    $$
    in a neighborhood of ##t=0## and investigate their behavior with respect to monotony (where defined). It is sufficient to determine the functions up to a differential equation. It's a mathematical problem, so existence will do.

    4. (solved by @DeathByKugelBlitz ) Find a ten digits integer with all ten digits, such that the first ##n## digits (##1\leq n \leq 10##) are divisible by ##n##. The integer should thus contain all ##10## different symbols ##0,1,2,3,4,5,6,7,8,9##.

    5. Is there always a position on a floor (continuous, no holes, steps etc.) for a rectangular table with four equal legs, such that the table does not wiggle?

    6. (solved by @SSequence ) Show that ##(0,1) \subset \mathbb{R}## cannot be written as a countable disjoint union of closed intervals.

    7. Areas and Volumes
    1. (solved by @Delta2 ) Show that the paraboloid
      $$P=\{\,(x,y,z)^\tau\in \mathbb{R}^3\,|\,x^2+y^2=z\, , \,x,y\in [-1,1]\,\} \subseteq \mathbb{R}^3$$
      and the hyperboloid
      $$H=\{\,(x,y,z)^\tau\in \mathbb{R}^3\,|\,x^2-y^2=z\, , \,x,y\in [-1,1]\,\} \subseteq \mathbb{R}^3$$
      have equal areas.

    2. Bring ##M =\{\,(x,y,z)^\tau\in \mathbb{R}^3\,|\,x^2\leq y^4\leq z^8\leq 1\,\}## into a normal form and calculate its volume.
    8. (solved by @Young physicist ) The table cards at a rotatable round table with 12 seats are set up for expected 12 guests. However, the persons ignore the cards and randomly distribute themselves to the seats.
    Is it always possible with a single turn of the table to make sure that at least two people sit in front of their correct table cards?

    9. Let ##p(x)=x^n+a_{n-1}x^{n-1}+\ldots +a_1x+a_0 \in \mathbb{R}[x]## be a polynomial where all roots are negative. Prove that $$\int_{1}^{\infty}\dfrac{1}{p(x)}\,dx$$ converges absolutely if and only if ##n>1\,.##

    10. (solved by @Young physicist ) Tom bets John that he can do the following: John will recite ##99## different numbers in the range ##1 \ldots 100## (i.e. positive integers) in a random order and he will be able to find and name the only number in that range that John will have missed. Tom will do it without taking any notes. What is the best algorithmic strategy to use in order to accomplish this? Give the relevant math.

    11. Let ##\emptyset \neq U \subseteq \mathbb{R}^+## be an open set, and ##x_0\in U\,.## We define the quotient logarithm of a function ##f\; : \;U \longrightarrow \mathbb{R}^+## at ##x=x_0## by
    $$
    f\,^-(x_0) := \lim_{x \to x_0} \dfrac{\log f(x)-\log f(x_0)}{\log x-\log x_0}
    $$
    Solve the 'differential equation' ##f^-=f\,.##

    12. Let [itex]x_0=0,x_1=1[/itex], and [itex]x_{n+1}=n(x_n+x_{n-1})[/itex] for [itex]n\geq 1[/itex]. Find [itex]\lim_{n\to\infty}\dfrac{x_n}{n!} \quad[/itex].

    13. Let ##p>q## be prime numbers such that ##p \not\equiv 1 \operatorname{mod}q##.
    Prove that each group with ##pq## elements is cyclic.

    14. If we multiply our five digits number by four, we will get the same number in reverse order. What's the number?

    15.

    16.

    17.

    18.

    19.

    20.

    21.

    22.

    23.

    24.

    25.
     
    Last edited: Dec 13, 2018 at 8:01 PM
  2. jcsd
  3. Dec 1, 2018 #2

    julian

    User Avatar
    Gold Member

    Problem 1:

    1)

    The partial sums

    ##
    s_m = \sum_{n=1}^m a_n
    ##

    are monotonic increasing as the ##a_n## are positive numbers. To show convergence we use the monotonic convergence theorem. "Every bounded monotonic sequence converges." (on this see for example the book "Introduction to metric and topological spaces" by W. A, Sutherland).

    If

    ##
    b_{n-1} {a_{n-1} \over a_n} - b_n \geq C
    ##

    for ##C > 0## and for all ##n > N## then

    ##
    b_N a_N - b_{N+1} a_{N+1} \geq C a_{N+1}
    ##
    ##
    b_{N+1} a_{N+1} - b_{N+2} a_{N+2} \geq C a_{N+2}
    ##
    ##
    b_{N+2} a_{N+2} - b_{N+3} a_{N+3} \geq C a_{N+3}
    ##
    ##
    \vdots
    ##
    ##
    b_{m-1} a_{m-1} - b_m a_m \geq C a_m
    ##

    Adding together gives and dividing by ##C## (recall ##C > 0##):

    ##
    \sum_{n=N+1}^m a_n \leq {b_N a_N \over C} - {b_m a_m \over C}
    ##

    Therefore

    ##
    s_m = \sum_{n=1}^N a_n + \sum_{n=N+1}^m a_n
    ##
    ##
    \leq \sum_{n=1}^N a_n + {b_N a_N \over C} - {a_m b_m \over C}
    ##
    ##
    < \sum_{n=1}^N a_n + {b_N a_N \over C}
    ##

    which is a constant not dependent on ##m##. Implying that ##s_m \equiv \sum_{n=1}^m a_n## converges.

    2)

    From

    ##
    b_{n-1} {a_{n-1} \over a_n} - b_n \leq 0 \quad \text{for all} \; n > N
    ##

    we have

    ##
    b_m a_m \geq b_{m-1} a_{m-1} \geq \cdots \geq b_N a_N \quad \text{for} \; m > N
    ##

    so that

    ##
    a_m \geq b_N a_N {1 \over b_m} \quad \text{for} \; m > N
    ##

    and hence

    ##
    \sum_{n=N+1}^\infty a_n \geq b_N a_N \sum_{n=N+1}^\infty {1 \over b_n}
    ##

    therefore if ##\sum_{n=1}^\infty b_n^{-1}## diverges then so does ##\sum_{n=1}^\infty a_n##.

    3)

    From

    ##
    b_n = \Big( n \cdot \Big( {a_n \over a_{n+1}} - 1 \Big) - 1 \Big) \log (n)
    ##

    rearranged we obtain

    ##
    n \log (n) {a_n \over a_{n+1}} - (n+1) \log (n+1)
    ##
    ##
    = b_n + \log (n) + n \log (n) - (n+1) \log (n+1)
    ##
    ##
    = b_n + \log (n) + n \log (n) - (n+1) \log \big(n (1 + {1 \over n} \big) \big)
    ##
    ##
    = b_n + \log (n) + n \log (n) - (n+1) \big( \log (n) + \log \big( 1 + {1 \over n} \big) \big)
    ##
    ##
    = b_n - (n+1) \log \big(1 + {1 \over n} \big)
    ##

    This allows us to employ parts 1) and 2). So that:

    ##
    \lim_{n \rightarrow \infty} \Big( n \log (n) {a_n \over a_{n+1}} - (n+1) \log (n+1) \Big)
    ##
    ##
    = \lim_{n \rightarrow \infty} b_n + \lim_{n \rightarrow \infty} - (n+1) \log \big( 1 + {1 \over n} \big)
    ##
    ##
    = \lim_{n \rightarrow \infty} b_n + \lim_{n \rightarrow \infty} - (n+1) \Big( {1 \over n} - {1 \over 2n^2} + {1 \over 3 n^3} - \dots \Big)
    ##
    ##
    = B - 1 .
    ##

    For ##B > 1## the right hand side is a number greater than ##0## and hence ##\sum_{n=1}^\infty a_n## converges.

    For ##B < 1## the right hand side is less than ##0##. Also, it is known that ##\sum_{n=4}^\infty (n \log (n))^{-1}## divergences. Hence ##\sum_{n=1}^\infty a_n## diverges.

    Here is a proof that ##\sum_{n=4}^\infty (n \log (n))^{-1}## divergences. We have that

    ##
    \sum_{n=4}^\infty {1 \over n \log (n)} \geq \int_4^\infty {dx \over x \log (x)}
    ##

    as ##1/x \log(x)## is monotonic decreasing function. Now

    ##
    {d \over dx} (\log (\log (x))) = {1 \over x \log (x)}
    ##

    as such

    ##
    \sum_{n=4}^\infty {1 \over n \log (n)} \geq \lim_{N \rightarrow \infty} \int_4^{N+1} {dx \over x \log (x)}
    ##
    ##
    = \lim_{N \rightarrow \infty} \big[ \log (\log (x)) \big]_4^{N+1}
    ##
    ##
    = \lim_{N \rightarrow \infty} \big( \log (\log (N+1)) - \log (\log(4)) \big)\rightarrow \infty .
    ##
     
    Last edited: Dec 1, 2018
  4. Dec 1, 2018 #3

    fresh_42

    User Avatar
    2017 Award

    Staff: Mentor

    Very nice, although I had hoped the third part to be more difficult.

    For all who are interested: part 1 and 2 are Kummer's criterion, part 3 is Bertrand's criterion.
     
  5. Dec 6, 2018 #4

    DeathByKugelBlitz

    User Avatar
    Gold Member

    Question 2, what is the date? 02/12/2018? :smile:
     
  6. Dec 6, 2018 #5

    fresh_42

    User Avatar
    2017 Award

    Staff: Mentor

    Damn, I should have postponed the question by eleven days. :wink:
     
  7. Dec 6, 2018 #6
    So, , for question 2, the actual date is 13/12/2018? Or the date is unknown to the reader?
     
  8. Dec 6, 2018 #7

    fresh_42

    User Avatar
    2017 Award

    Staff: Mentor

    The date here is only the day, and yes, the actual date is not needed.
     
  9. Dec 6, 2018 #8
    Thanks!!!

    Problem 2:

    36 can be fatorated into 2*2*3*3 = 36. Therefore, the possibilities are:
    4, 3, 3 , day 10
    6, 2, 3, day 11
    9, 2, 2 day 13

    There is when my English starts to itch my head on something: if that son is the "eldest" and not the "elder", does that mean he has another son who is also older than another son? If that's the case, if we consider "born at the same day" as "same age", they can't be twins.

    So their ages are 6, 2 and 3.

    (Quite embarassed if that whole thing I came up with the word "eldest" actually made me fail the question instead of solving it ahahaha)
     
  10. Dec 6, 2018 #9

    fresh_42

    User Avatar
    2017 Award

    Staff: Mentor

    He has an oldest son. That's the information.
    Your result isn't the solution. You skipped an information which is a bit hidden in the words.
     
  11. Dec 6, 2018 #10

    fresh_42

    User Avatar
    2017 Award

    Staff: Mentor

    ##3\cdot 9\cdot 1 \neq 36## and you don't know anything about ##13##, at least not in an acceptable solution. In fact the date is only needed to exclude ##1\cdot 1 \cdot 36## as its sum would be ##38##, which is not a date. You can only use what is given by the original text. Don't skip an information!
     
  12. Dec 6, 2018 #11
    Problem 2a:

    Given that there are 3 sons and the product of their ages is 36, the possible combinations of sons' ages are:

    (1, 1, 36)
    (1, 2, 18)
    (1, 3, 12)
    (1, 4, 9)
    (1, 6, 6)
    (2, 2, 9)
    (2, 3, 6)
    (3, 3, 4).

    Note that the second mathematician did not have the solution once the first stated that the ages summed up to the date. This means that there is more than one combination of ages that sums to the date. Only two of these combinations result in the same sum:

    (1, 6, 6) and (2, 2, 9).

    Finally, the first mathematician reveals that he has one eldest son, which rules out (1, 6, 6).

    Thus, the ages of the mathematician's sons are 2, 2, and 9.
     
  13. Dec 6, 2018 #12

    DeathByKugelBlitz

    User Avatar
    Gold Member

    4:

    When n = 10, the number has to end with a zero, otherwise it's not divisible by 10.

    I can start with 1, 2, 3, 4, 5, 6, 7, 8, or 9 (all are divisible by 1).

    I can have a possible 50 numbers that fit the criteria for n=2 (all even 2 digit numbers). Since when n = 4, 6, 8 we need the number to be divisible by 4, 6, 8 respectively, and these numbers are divisible by 2, the 4th, 6th and 8th digit must be even.

    For n = 5, the nth digit must be either 5 or 0

    So when n = 10, the nth digit is 0
    And for n = 2, 4, 6, and 8, we know that the nth digit is either 2, 4, 6, 8, or 0
    For n = 5, the nth digit is 5 or 0

    Will continue trying to figure the rest out, that's what I have so far :smile:
     
  14. Dec 7, 2018 at 7:53 AM #13

    DeathByKugelBlitz

    User Avatar
    Gold Member

    Using the above, and then just testing numbers (brute force) I reached 1,020,005,640
     
  15. Dec 7, 2018 at 8:01 AM #14

    fresh_42

    User Avatar
    2017 Award

    Staff: Mentor

    Sorry that was my fault (and I will correct it immediately) and beg you a pardon, because I have been unclear. I meant that all 10 digits should actually occur as in 1234567890.
     
  16. Dec 7, 2018 at 8:02 AM #15

    DeathByKugelBlitz

    User Avatar
    Gold Member

    I thought that would be the case, so each unique digit can only be used once :)
     
  17. Dec 7, 2018 at 8:09 AM #16

    fresh_42

    User Avatar
    2017 Award

    Staff: Mentor

    Yes. E.g. the last one has to be 0 and the second one has to be even, etc.
     
  18. Dec 7, 2018 at 8:50 AM #17

    DeathByKugelBlitz

    User Avatar
    Gold Member

    When n = 10, the number has to end with a zero, otherwise it's not divisible by 10.

    I can start with 1, 2, 3, 4, 5, 6, 7, 8, or 9 (all are divisible by 1).

    I can have a possible 45 numbers that fit the criteria for n=2 (all even 2 digit numbers). Since when n = 4, 6, 8 we need the number to be divisible by 4, 6, 8 respectively, and these numbers are divisible by 2, the 4th, 6th and 8th digit must be even. However, the first digit must be odd, since all the even digits are now taken.

    For n = 5, the nth digit must be either 5 or 0. But 0 is taken for n = 10, so must be 5.

    So when n = 10, the nth digit is 0
    And for n = 2, 4, 6, and 8, we know that the nth digit is either 2, 4, 6, or 8.
    Therefore, we also know that for when n = 1, 3, 7, 9 the nth digit is either 1, 3, 7, 9
    For n = 2, 6 nth digit is either 4 or 8
    For n = 4, 8 nth digit is either 2 or 6
    For n = 5, the nth digit is 5

    all the allowed 2 digit numbers:

    14, 18, 34, 38, 74, 78, 94, 98

    For each of these we add an unused odd number to the end and see if the result is divisible by 3

    147, 183, 189, 381, 387, 741, 783, 789, 981, 987

    All possible 5 digit numbers (all possible 4 digit numbers, plus a 5 at n=5 since the 5 has to be there):

    14725, 14765, 18325, 18365, 18925, 18965, 38125, 38165, 38725, 38765, 74125, 74165, 78325, 78365, 78925, 78965, 98125, 98165

    All possible 6 digit numbers (if 2nd digit is 4 or 8, 6th digit must be 8 or 4), then checking if they are divisible by 6

    147258, 183654, 189654, 381654, 387654, 741258, 783654, 789654, 981654

    All possible 7 digit numbers:

    1472583, 3816547, 7836549

    All possible 8 digit numbers:

    38165472

    There is only 1 possible 8 digit number

    The number after 9 digits becomes 381654729

    And the 10 digit number is 3816547290
     
  19. Dec 7, 2018 at 9:08 AM #18

    DeathByKugelBlitz

    User Avatar
    Gold Member

    It was fun although a bit brute force :-p
     
  20. Dec 8, 2018 at 11:05 AM #19

    fresh_42

    User Avatar
    2017 Award

    Staff: Mentor

    Fun is important. Brute force? Maybe, but the exclusion principle is probably the by far most used thought in mathematics: "##P(x)## true. Imagine ##P(x)## fasle, then ..." and so all possibilities with a false property ##P(x)## are excluded. But it is also used directly, when solutions are narrowed down, e.g. in numerical analysis, whenever an initial condition in a differential equation system is applied, because it excludes all other trajectories, and probably a lot more.
     
  21. Dec 8, 2018 at 11:18 AM #20

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Re problem 2a:

    Although, technically, he could have two sons of the same age that are not twins! Sorry, that's a real spoiler!
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted