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2014 Sun Life Financial Canadian Open Math Challenge - Nov 6/7

2014 Sun Life Financial Canadian Open Math Challenge - Nov 6/7


Prof. Robert Woodrow

Problem of the Week

by Prof. Robert Woodrow, COMC 2014 chairman

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.

Because past COMC exams are available on the CMS website, in order to give you a further taste of the different types of challenge you may find we present some problems drawn from other contests in Canada and from around the world. I hope they will be fun and help polish your skills for the COMC this November. Solutions will be presented the following week so you can compare your solution and reasoning with the posted solution.

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.

Week 3

  • Problem (posted September 16th)

    This week we look at a problem which draws on basic algebra from high-school.

    Let $x,y,$ and $z$ be real numbers such that $\begin{array}{ccc} x+y+z =3 & \mbox{and} & xy + yz + xz = a \end{array}$, where $a$ is a real parameter. Determine the value of the parameter $a$ for which the difference between the maximum and the minimum possible values of $x$ equals $8.$

  • Solution (check back on September 23rd)

Week 2

  • Problem (posted September 9th)

    The problem for this week is a problem involving velocities. It takes some thought to find how to get started!

    Two cars 100 metres apart are traveling in the same direction along a highway at the speed limit of 60 km/h. At one point on the highway, the speed limit increases to 80 km/h. Then a little later, it increases to 100 km/h. Still later, it increases to 120 km/h. Whenever a car passes a point where the speed limit increases, it instantaneously increases its speed to the new speed limit. When both cars are traveling at 120 km/h, how far apart are they?

  • Solution (posted September 16th)

    This was problem 3 of Part 2 of the 2013-2014 Alberta High School Prize Examination.

    Let the first car be at a point $B$ while the second car is at a point $A$ , both in the 60 km/h zone. Then $AB = 100$ meters. Let the first car be at a point $D$ while the second car is at a point $C$, both in the 120 km/h zone. Now the amount of time the second car takes to go from $A$ to $C$ is the same as the amount of time the first car takes to go from $B$ to $D$. Both cars take the same amount of time going from $B$ to $C$ . Hence the amount of time the second car takes to go from $A$ to $B$ at 60 kph is the same as the amount of time the first car takes to go from $C$ to $D$ at 120 kph. It follows that $CD = 2AB = 200$ meters.

Week 1

  • Problem (posted September 2nd)

    For the first week we look at two problems. The first does not require work to be shown, and is a puzzle problem. The second requires work to be shown. It involves digits of numbers and divisibility. Look next week to check your solution.

    Problem A
    A home has some fish, some birds and some cats. Altogether there are 15 heads and 14 legs. If the home has more than one of each animal, how many fish are there?

    Problem B
    There are 2014 digits in a row. Any two consecutive digits form a number which is divisible by 17 or 23.

    1. If the last digit is 1, then what are the possibilities for the first digit?
    2. If the first digit is 9, then what are the possibilities for the last digit?
  • Solution (posted September 9th)

    Problem A
    [This problem was A6 from the 2014 Calgary Junior Mathematics Contest. ]
    Let there be $F$ fish, $B$ birds and $C$ cats. From the legs we get $2B +4C =14$, so $B + 2C =7$ and from the heads $F+B+C = 15$. Now there are more than one of each type, so there are at least 2 birds. But then there can be at most 2 cats, so $C=2$ and $B=3$. From this we get $F = 15-2-3 =10.$

    Problem B
    [This problem was B2 from the 2014 Calgary Junior Mathematics Contest. ]

    1. The two digit multiples of $17$ are $17,\ 34, \ 51, \ 68$ and $85.$ Those of $23$ are $23,\ 46,\ 69$ and $92.$ We now can work backward from right to left working two digits at a time starting with $x1$. This gives $51$, then with $x5$ we get $851$, then $6851$, then $46851$, $346851$, $2346851$, $92346851$, and next $692346851$. It is now clear that the pattern $92346$ of length five will repeat. Now $2014 = 402 \times 5 + 4$, so the number must be $402$ blocks of $69234$ followed by $6851$, so the first digit is $6.$
    2. With the same method as before but starting with $9y$ we get $92$, then $923$, then $9234$ and then $92346$. At this point we have both a multiple of $17$, namely $68$ and a multiple of $23$, namely $69$ to consider. Let us first pursue the path with $68$, i.e. continuing to $923468$, then $9234685$, and $92346851$ and next $923468517$, where we are stuck since we cannot have $7y$ appearing. What happens with the possibility from $69?$ Here we observe, as in the first example that a cycle $92346$ of length $5$ is formed. Now we see that two possible numbers could be formed. One is $402$ blocks of $92346$ ended by $8517$ and the other is $402$ blocks of $92346$ ended by $9234$. So the two possibilities for the last digit are $4$ and $7.$

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


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