It’s about the singular episodes of the 35 camels that were to be divided btwn three brothers. How Beremiz made an apparently impossible division that leads the brothers completely satisfied.
the Story Rewritten (Simplified)
Close to an old, half-abandoned inn, there were three men arguing heatedly beside a herd of camels. Amid shouts these men gestured widely, and we heard:
“It cannot be!”
“That’s not fair!”
My friend Beremiz the Man who counted, then approach them and asked what happened.
The oldest said: “we are brothers, and have received these 35 camels as inheritance. My father has left a will to give half (1/2) herd to me, one-third (1/2) to my younger brother and (1/9) to the youngest. But difficulty find us. If half herd is 17 (1/2), and one-third is not precisely an integer, them how can we make the decision?”
The herd was divided to the brother’s satisfaction. While Berimiz travelled with me on one camel, we left the brothers with two camels: Berimiz and I each ride one.
Dear readers, it’s your turn to find the solution – do it!
Here is a hint:
The brothers are supposed to get livestock of camels, so camels distributed to each brother has to be an integer. If we start with a herd of camels which is a common multiple of 2, 3, and 9, then there would be no trouble at all ! — By the way, what’s that number?
We shall always try alternatives! It’s fun, and it expands our horizon in thinking.
None of the numbers 2, 3, 9 can divide 35. So if we follow the father’s will, each brother get a fraction. How can we please everyone? If we give each brother more than the portion according to father’s will, for sure he shall not complain.
Fortunately, in the context of this question, it’s possible to do this!
What is this? It’s a collection of math adventures! Here is the cover.
For some who have not yet read this book, it’s rightful to be surprised at the title. Is it true that everyone can count? Yes but don’t take it for granted.
The hero of this book, Beremiz, is so talented at math that there is no problem he can not solve – even those that seems impossible. He knows certainly much more than “counting”! Let’s follow his adventure and follow his ideas to learn the smart way of thinking!
Malba Tahan is the author of the book, and also the traveller with Beremiz in the book.
What are common divisors and common multiples? Let’s learn this together with an example. Given two numbers 12 and 9, each has its divisors as:
Example: List the common divisors and common multiples of 12 and 9.
Divisors of 12: 1, 2, 3, 4, 6, 12.
Divisors of 9: 1, 3, 9.
Let integers N>0, M>0. Recall quickly that a divisor of N is a positive integer (counting number) d<N such that d divides into N evenly (that is, remainder is zero). Given N and M, a common divisor is one divisor of N and of M both.
In the example above, the common divisors of 12 and 9 are 1 and 3, while 1 is the least, and 3 is the greatest – usually written as GCD (Greatest Common Divisor) or GCF (Greatest Common Factor).
Now let’s turn our attention to the common multiples.
Multiples of 12: 12, 24, 36, 48, 60, 72, ..
Multiples of 9: 9, 18, 27, 36, 45, 54, 63, ..
Recall quickly a multiple of N can be obtained by N multiplied by any positive integer. Given N and M, a common multiple of N and M, as said in its name, is the common multiple of N and of M both.
In the example above, the common multiples of 12 and 9 are 36, 72, etc. 36 is the least one, usually referred as LCM (Least Common Multiple).
This discussion also gives us one way of finding GCF and LCM: just list common factors /divisors of each and common multiples of each, then circle the greatest of common factors and the least of common multiples. This way, of course, stick strictly to the definition.
Is it a better way to find the GCF and LCM of two numbers?
Yes, indeed. We know that 9 = 3 x 3. Let’s check whether 12 can be divided by 3 (Yes) and by 9 (No). We can conclude that GCF of the two numbers is 3.
How about LCM? Let us divide 12 by 3 (which is the GCF) to get 4, and multiplies by the other number 9, we get 36 – which is the LCM. Or we can divide 9 by the GCF 3, and multiplies by the other 12, again we get 12 the LCM.
Wow! This seems much easier. Why this approach works?
We’d better do some investigation to find out – but before doing that, let’s verify by another example.
Find the GCF and LCM of numbers 108 and 40.
Ohh, GCF is 4 and LCM is 1080. Have you guessed it correct?
One hint for why the approach works is, in the above two examples, if you multiply the GCF and LCM, you will find the product to equal exactly the product of the two given numbers, as:
9 x 12 = 3 [GCF] x 36 [LCM], and 40 x 108 = 4 [GCF] x LCM [LCM].
Based on this there are also games (GCF-LCM Web), would you like to have a try?
Another interesting problem in Galois 2017 is the following one, which considers the Koeller Shading Ratio. The problem is as below.
Problem. A Koeller rectangle:
is an m by n rectangle where m,n are integers with m>=3 and n>=3,
has lines drawn parallel to its sides to divide it into 1 by 1 squares, and
has the 1 by 1 squares along its sides unshaded and the 1 by 1 squares that do not touch its sides shaded (as in figure).
An example of a Koeller rectangle with m = 6 and n = 4 is shown. (The cell marked S is shaded, the blank cell is unshaded.)
For a given Koeller rectangle, let r be the ratio of the shaded area to the unshaded area.
Now are the questions:
(a) [Answer only] Determine the value of r for a Koeller-rectangle with m = 14 and n = 10.
(b) [Full solution required] Determine all possible positive integer values of u for which there exists a Koeller-rectangle with n = 4 and r = (u/77).
Well, readers are advised to try first before reading the solution below, followed by some remarks.
First, develop some formulae.
Unshaded region = m + m + n + n – 4 = 2 (n+m-2)
Shaded region = (m-2) (n-2) = mn – 2m – 2n +4
Ratio r (of shaded to unshaded)
r = (m-2) (n-2) ⁄ (2 (n+m-2))
For (a), m = 14, n = 10, therefore
r = (12) (8) ⁄ (2 (22)) = (24) ⁄ (11) .
For (b), since n = 4, r = (m-2) ⁄ (m+2)
As its said that r = (u / 77), so:
u = 77 r = (77) (m-2) ⁄ (m+2) (*)
As (m-2) ⁄ (m+2) = 1 – [4 ⁄ (m+2)], only 2 or 4 divides both m+2 and 4 (besides 1 as trivial factor).
In order that u be an integer, m+2 must contains a factor of 77, as well a factor 2 or 4.
As 77 = 7 x 11, so we will consider the cases
(a) m+2 = 7, 14, 28 –> m = 5, u = 33; m = 12, u = 55; m = 26, u = 66;
(b) m+2 = 11, 22, 44 –> m = 9, u = 49; m = 20, u = 63; m = 42, u = 70;
(c) m+2 = 77, 154, 308 –> m = 75, u = 73; m = 152, u = 75; m = 306, u = 76.
Therefore, possible values of u are 33, 55, 66, 49, 63, 70, 73, 75, 76 (total nine solutions).
(1) For Koeller shading ratio r, we recommend to work with formula from the very beginning; it is clearer and reduce repetitious work. The formula links r with n, m.
(2) In question (b), given 77 as the common denominator of a series of ratios, it asks to find all possible integers u to appear in the numerator – this requires an analysis of factors – as in reduction of fractions.
(3) As u is an integer, an explicit expression of u – as in (*) — is convenient to work with.
Finally, note that has 3 x 2 x 2 = 12 factors. But m+2 cannot be 1, 2 or 4 (as m>=3). —
Therefore, the other 12 – 3 = 9 factors correspond to the nine solutions.
This is just a step-by-step work with algebra, plus some analysis of factors. A neat and clear presentation is what is asked by the problem.
In this year’s Galois math contest, one problem about the special-pricing day breakfast is fairly interesting. Here we take questions (b) and (d) of this problem to the readers for a glimpse into, – and even to taste its flavour.
The Breakfast Restaurant has a special pricing day. If a customer arrives at the restaurant between 4:30 a.m. and 7:00 a.m., the time that they arrive in hours and minutes becomes the price that they pay in dollars and cents.
For example, if a customer arrives at 5:23 a.m., they will pay $5.23.
(b) [Answer only] Robert arrived 10 minutes before Emily, and both arrive during the period of the special pricing. In total, they paid $12.34. What were their arrival times?
(d) [Full Solution Required] Larry and Mio arrived separately during the period of the special pricing. In total, they paid $11.98. Determine the range of times during which Larry could have arrived.
The problem is so easy as everyone can understand it, right? – But it takes some thinking to get the solution correct and perfect! By the way, when it’s suggested “Full solution required”, students will have to write not only the final answer, but also solution steps.
If we skim forward to questions (b) and (d), we find a group of two persons come to the restaurant, and it’s given of the price they paid together; and then students (writing the test) is asked to deduct about amount of each paid as well as arriving times according to some more given conditions. Not in a hurry for details, one may ask self (as a warm-up) that what total price paid is possible (valid)?
The price for two people must be between $8.60 (if both arrived at 4:30 am) and $14.00. But some price for example, $13.30 is not possible. Suppose both arrived at 6: 59, they paid $6.59 each, so total of $13.18; if one of them arrived (just one minute late) at 7:00, the total paid shall became $13.59. If both arrived $7.00, that’s $14.00 total. – Obviously a jump! It’s caused by 1 hour = 60 min however $1 = 100 cents.
For question (b) it’s straight and we can separate hours with minutes. The hour they arrived? Just divide 12 by 2 to get 6. What’s the minutes they arrived? It’s like finding two integers with sum 34 and difference 10. You got it? Robert arrived at 6:12 am and Emily did 10 min later at 6:22.
Any other possibilities? Well no. We know we will convert 100 cents to $1. With the total price given as 12.34 cents, is it possible that the cents they should pay (according to the minutes they arrive) add up to 134 cents? Not possible, since the maximum is 59 + 59 = 118 (We’ve done that!). — The thinking does not yield new solution but as many know, contest questions may be tricky so be wary.
For question (d), the total cents paid were 98 cents (NOT possible to be 198 cents), so we are sure completely that the hours they arrived add to 11 (= 5+6), and the minutes they arrived add to 98. Decomposing 98 as the sum of two integers, while both being between 0 and 59, we have:
98 = 39 + 59 = 40 + 58 = … = 59 + 39
But here indeed there’s a tricky stuff: it’s said in the question that Larry and Mio arrived separately, no mentioning about who were earlier.
The one arrived early was after 5:00, and the late comer was after 6:00. Therefore, Larry could have arrived between 5:39 to 5:59 or between 6:39 to 6:59.
Worthy to note that it’s not possible for any of the two to arrive between 6:00 to 6:38.
With some additional effort, one shall be able to present these steps of solution nicely.
Galois /Hypatia contests are Canada-wide math competitions (for grade 10 resp. grade 11). The contest writing date for 2017 is on April 12th Wednesday.
We, Jonah’s Math Corner, are now planning a short course for warming-up and preparing for these contests. It is planned to have 6 classes each lasting 2 hours (so the total time will be 12 hours), starting from the last week of February and continue until the first week of April (with one week off for the spring break at end of March). There is one series for each of the contests. The preparation series for Galois (grade 10) contest is tentatively scheduled to be on Thursday evening, and the preparation series for Hypatia (grade 11) is scheduled on Saturday morning.
Prepare for Galois contest: February 23 – April 6th, every Thursday evening (call for details)
Prepare for Hypatia contest: February 25 – April 8th, every Saturday morning (call for details)
Contest writing date: April 12th 2017
Place: (call for details)
Fee: $270 per student, due on first class.
Please contact us now before the class starts — we reserve the rights to cancel the class if no sufficient interest.
We encourage friends to spread the news to those who are interested in taking the challenge! [Applause] [Rose] [Rose] [Smile]
The students may choose to write the contests either at Jonah’s Math Corner, or at the local school he/she attend (since some public schools of Calgary will also supervise the contest). The fee for contests are extra ($22 if choosing to write at Jonah’s Math Corner).
For interested students or their concerned parents, please contact by calling 403-690-8878 or email email@example.com. We will provide detailed information. If you are interested but feel the schedules time does not fit in your busy schedule, please let us know – we are always ready to make some adaptations.
We guide and train students in problem-solving. For two consecutive years (2015 and 2016), students from Jonah’s Math Corner performed well and have come out top nationwide* in Fryer (grade 9) and Galois contest (grade 10).
* Please note that the past performance from Jonah’s math corner should not be interpreted as an indicator for current students’ success for oncoming contests. There is no guarantee in any form on students results after taking the preparation series for the contests.
About Galois /Hypatia contests
Galois /Hypatia contests are Canada-wide math competitions (for grade 10 respectively grade 11).
Centre of Education in Mathematics and Computer Science, within University of Waterloo, organizes and coordinates the contests. The top performers will be awarded certificates of excellence and be recognized at the local school and on CEMC website.
Galois /Hypatia contests have answer-only problems (requiring a student to work out his/her own answer) as well as full-solution problems that require full presentation of solution steps.
The problems in these two contests match closely with Canadian math curriculum across provinces of Canada. The problems challenge students to show well-trained math talents and math expression skills while still focusing on the fundamental of school math curriculum.
When we study the shapes, we often mention “geometric properties” — do not be scared by the words “properties”, its just some sure-things about a shape, like “circle is round”.
This is a fairly simple graph about the shapes.As shown, ABDE is the square; C is its centre, and the circle centred at C passes through all the four corners: points A, B, D and E.
Here are a number of properties about the square, and its circum-circle: (the circle passing through every corner of the square)
1) The four sides of a square are congruent (equal in length).
2) Each of the interior angle (at the corner) of the square is a right angle, thus these interior angles are congruent.
3) Each square has a centre, which is the intersection point of the two diagonals.
4) The two medians (traversal of the opposite sides) intersect at the centre.
5) The two diagonals of a square bisects each other.
6) Each diagonal divides the square into two regions with equal area, and the two diagonals divide the square into four equal-area regions.
7) Each square is inscribed into a circle, with the centre of circle to be located at centre of the square.
8) For any square, when rotating by 90 degrees, new square will be coincident (overlapping exactly) to the old square. This is true when rotating by any multiples of 90 degrees (since we can apply rotation again and again).
Using a compass, you can draw a circle at any place, with any radius.
Now let’s reverse the problem. Given a circle, do you know how to find its center? (Of course, once the circle is found, there shall be no problem at all to tell its diameter, or radius.) You only see the circle itself, there is no explicit indication on where the center is.
With two set of restrictions on what kind of tools you can use, there are actually two questions. In general, you can find the center using any convenient method, including copy-and-paste the circle onto a paper, and then fold it. In particular (from classical Euclidean geometry), where it’s required to do so with a ruler (with which you are allowed to draw lines and line segments only) and a compass (with which you are allowed to draw circles only).
See the following article on how to do it in general.