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The 2019 Canadian Open Mathematics Challenge -- Nov 7/8

The 2019 Canadian Open Mathematics Challenge — Nov 7/8


Nicolae Strungaru

Problem of the Week

by Prof. Nicolae Strungaru, Grant MacEwan University

The COMC has three parts. In part A solutions do not require work be shown and may be possible to do in your head. In part B the problems begin to draw on more knowledge and have some more challenging aspects that will need a pencil and paper to solve. By Part C the problems require that work be shown and involve arguments to support the answer.

We have selected a cross section of contest problems from a variety of national and regional contests that we hope will stimulate interest in problem solving and give some experience to get ready to write the Canadian Open Mathematics Challenge in November. The problem areas are not tied to particular grade levels, or to the curriculum but cover a number of areas from algebra, through logic and some geometry.

We will post solutions to these problems one week later, but teachers should be aware that determined students may be able to locate solutions elsewhere online before then.

For a more comprehensive set of problems and solutions at each of these levels, please feel welcome to download past official exams and solutions from our archive.

Need more COMC Problems-of-the-Week? Take a look at the set from past years: 2018, 2017, 2016, 2015, 2014, or 2013!

Week 2

  • Problem (posted September 10th)

    How many subsets with three elements can be formed from the set $\{ 1 , 2 , 3, \dots , 20 \}$ such that $4$ is a factor of the product of the three numbers in the subset?

  • Solution (check back on September 17th)

Week 1

  • Problem (posted September 3rd)

    We give two entry level problems this week. Give them a try. Look for the source and the solution next week!

    Problem A
    Let $x$ be an integer of the form $$ x=\underbrace{111\dots11}_{n \mbox{ times }} $$ Show that, if $x$ is prime then $n$ is prime.

    Problem B
    Given a rectangle with sides $a$ and $b$, with $a>b$, fold it along the diagonal. Determine the area of the overlapping triangle (the shaded area in the picture).

  • Solution (posted September 10th)

    Problem A

    Problem 7 of the Chilean Mathematical Olympiads 1994-95, which appeared in Crux Mathematicorum [2001: 170]. We present the solution by Christopher J. Bradley that appeared at [2003: 294].

    Note: A video solution to this problem is also available.
    Note that $$ 9x=\underbrace{999\dots 99}_{n \mbox{ times }}=10^n-1 \,. $$ Therefore, $$ x=\frac{10^n-1}{9} \,. $$ Now, if we assume by contradiction that $n$ is composite, $n=n_1n_2$ for some $n_1, n_2 >1$. Then $$ x=\frac{10^n-1}{9}=\frac{10^{n_1}-1}{9} \left(1+10^{n_1}+10^{2n_1}+\dots+10^{n_1(n_2-1)} \right)\,. $$ As $\frac{10^{n_1}-1}{9}$ is an integer greater than $1$ and $1+10^{n_1}+10^{2n_1}+\dots+10^{n_1(n_2-1)}>1$ we get that $x$ is composite, a contradiction.

    Since, we got a contradiction, our assumption that $n$ is composite is wrong.

    Therefore, $n$ is not composite and $n\neq 1$ (since $x=1$ is not prime), which proves that $n$ is prime.

    Editor's note: Primes of this form are sometimes called "repunit primes". So far 5 repunit primes are known, namely the values of $x$ for $n \in \{ 2, 19, 23, 317, 1031 \}$. The values of $x$ for $n \in \{ 49081, 86453, 109297, 270343\}$ are also believed to be prime.

    Problem B

    Problem 1 of the XV Gara Nazionale di Matematica 1999, which appeared in Crux Mathematicorum [2002: 481] . We present the solution by Bob Serkey that appeared at [2005: 37].

    Note: A video solution to this problem is also available.

    Let $E$ be the intersection of $BC$ and $AD$, and let $x=BE$.

    Note that the right triangles $\Delta AEB$ and $\Delta CED$ have $CD=AB$ and $\angle AEB = \angle CED$, and thus are congruent. Therefore, we have $$ DE=BE=x \,. $$ We therefore also have $$ AE=CE=a-x \,. $$ By Pythagoras theorem we have $$ (a-x)^2+b^2=x^2 \, \Rightarrow \, x=\frac{a^2+b^2}{2a} \,. $$ In the triangle $\Delta BED$, $DC=b$ is the altitude to the base $BE=x$. Therefore $$ \mbox{Area} (BED)=\frac{bx}{2}= \frac{b}{4a} \left(a^2+b^2 \right) \,. $$

To report errors or omissions for this page, please contact us at comc@cms.math.ca.


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