## Math Story: Dividing the Herd of Camels

3rd Story

Dividing the Herd of Camels

Summary:

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 Conclusion:

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.

Solution:

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?

Alternative Solution:

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!

## Adventure with Beremiz – the Man Who Counted

The Man Who Counted

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.

You can continue to start the amazing story.

## GCF-LCM Stuff: Find them: how, why, find smart and fast

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?

## Guide to Contents

The latest posts discuss problems in the recent Fryer / Galois Contest (for Grades 9&10 students). Those problems and their solution process are really interesting.

For similar topics, choose |* Windows of Math Contests *|.  If you are looking for other contents, we have an array of interesting topics. (Smile)

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## Preparation for Galois /Hypatia Math Contests – Jonah’s Math Corner offers a six-week short series

Jonah’s Math Corner Announcing:

Preparation for Galois /Hypatia Math Contests

– a six-week short series

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.

In summary:

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.

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 info@mathcorner15.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.

Past Results

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 click below for the Honour Roll –

the contest results for students at Jonah’s Math Corner

* 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.