## Tuesday, January 26, 2016

### [S1_20160126FMNT] Largest Common Remainder for Three Divisors

Problem / Question

 Find the greatest 4 digit number that will divide 63, 45 and 69 so as to leave the same remainder.

Solution
LCM + (smallest number – 1) = 7245 + 44 = 7289   J

Remarks
This question is from National University of Singapore High School, which caters to students who are very interested in science and mathematics, and possibly an academic career.  Do not be fooled by my short and sweet solution.  The question is actually quite challenging, and I went by a long way before coming up with this elegant solution.  This reminds me of  Human Resource managers who think that more lines of code written by programmers means more work is done.  Actually, a lot of hard thinking could be involved in writing a one-line code that does the same job.
Research shows that when expert problem solvers realise that they are stuck, they change tactics.  They go back to the drawing board.  “Insanity is doing the same thing over and over again and expecting the different results.” said Albert Einstein.  The five stages of mathematical problem solving are
1. Understanding the problem
2. Planning a strategy
3. Executing the plan
4. Evaluation
5. Reflection (which can even include blogging about it!)
At the evaluation stage, if one finds that one is not getting the results, or if the approach is not elegant, one goes back to stage 2 to devise a new strategy.  I had tried using a complicated Chinese Remainder Theorem approach, got the solution after one page of work, and realised that the problem can be solved very simply.

How does the solution work
The set of remainders dividing by  63,  45  and  69  repeat themselves in a cycle and the length of the cycle happens to be the Lowest Common Multiple (LCM).  We can see this quite easily: suppose  x  and  y  both give a remainder  R  dividing by  63,  45  and  69,  then  xy  gives a remainder of  0  when divided by  63,  45  and  69,  which means  xy  is a common multiple of  63,  45  and  69, of which the lowest is the LCM.
It is a routine matter to get the LCM via prime factorisation as follows:   63 = 32 × 7,   45 = 32 × 5,  and   69 = 3 × 23.  We pick the highest power for each occurring prime and we obtain  LCM = 32 × 5 × 7 × 23 = 7245.
The next thing to note is that remainders must be less than the divisors.  Hence the largest common remainder must be  44,  one less than the smallest divisor.  You cannot have a remainder of  45  when divided by  45,  because you could simply have bumped up the quotient (the result of division) by 1 and get zero remainder.  So from 7245, 7246, 7247, ... to 7289,  you get common remainders of  0, 1, 2, ..., 44.  Once you hit  7290,  the remainder for division by  45  will hit  0  while the remainders for  63  and  69  will be  45  but this will not be a common remainder.  The next time we get a common remainder will be  2×LCM = 2×7245 = 14490  which is  5 digits long.  So within  4  digits, the highest number with a common remainder is  7289,  which gives a common remainder of  44.

H04. Look for pattern(s)
H05. Work backwards
H09. Restate the problem in another way
H10. Simplify the problem
H13* Use Equation / write a Mathematical Sentence

Suitable Levels
Lower Secondary Mathematics (Sec 1 ~ grade 7) challenge
* other syllabuses that involve factors and multiples or number theory
* any precocious or independent learner who loves a challenge

## Monday, January 25, 2016

### [AM_20160125DAHT] Horizontal Tangents via Quadratic Discriminants

Problem

Introduction
This is a Additional Mathematics textbook problem.  This question is of an intermediate level of difficulty.  The general method is by differentiation.  The equation of the curve happens to be capable of being put into a quadratic equation in  x.  Hence we can also use the theory of quadratic discriminants.  I present both methods of solution.

Method 1 (Using differential calculus)

Heuristics Used
H04. Look for pattern(s)
H05. Work backwards
H09. Restate the problem in another way
H11. Solve part of the problem
H12* Think of a related problem
H13* Use Equation / write a Mathematical Sentence

Suitable Levels
* other syllabuses that involve differentiation or quadratic discriminants
* any independent learner who is interested

## Friday, January 22, 2016

### [Maths_Education] The Need to Harness and Transcend Technology

Problem / Question

 Help me please!  My teacher needs it tomorrow! How many millilitres are there in 3.4 litres?

LOL!  Well, if she really needs it, ask her to type the question into Google!

Remarks
Actually, this is not a joke.  Many easy questions in school mathematics are now answerable by Google.  In fact, Wolfram Alpha is able to answer mathematically more difficult but routine questions.  Wikipedia and YouTube are also useful for learning mathematics.  There are many other good resources available on the Internet and public libraries.  The sad thing is,
 1) It seems that our current generation of kids views homework as a chore to be done for the teacher, not as an experience to be used for their own learning.  OK, maybe the homework task should have been designed better, to ask non-Googleable questions, but educators need to be aware of that is happening to our kids. 2) Our kids do not know how to choose and use the abundantly available technology and resources 3) Neither are they taught how to do this in school (do the teachers know it themselves?) 4) This type of question merely targets the lower levels of Blooms Taxonomy 5) Even for these easy questions, children are unable or unwilling to make the effort to find their own answers, and how are they to engage in Higher Order Thinking, creative thinking, reasoning etc.?
As technology evolves and improves and replaces many jobs, and as our children de-evolve and slacken, when this generation grows up they will not only face an even more challenging environment for their careers than today, they may not have the right values, attitudes and dispositions for living life.
The purpose of learning mathematics in school should be to learn how to think and to serve others.  Human students must be educated to use technology and go beyond technology, to seek answers and to help other people instead of merely relying from other people for “help”.  This is one of the reasons why an identity (“learning to be a type of person”) approach to learning mathematics is important.

## Tuesday, January 19, 2016

### [S1_Expository] Exploring the HCF and LCM with a calculator

How does it work?
When we reduce a fraction to its lowest terms, we actually cancel out as many common factors as possible.  So eventually the numerator and the denominator of original fraction get cancelled by their Highest Common Factor (HCF), a.k.a. “Greatest Common Divisor (GCD) ” in America.  Just as England and the United States are divided by a common language(*), we can divide  756  by  6  to get the HCF.  Why?  That is because  756  was divided by the HCF to get  6.  You may try a similar trick with the denominators.  You get the same conclusion.
It is useful to know that the product of two numbers is equal to the product of their  HCF  and  LCM.  So   756 × 1386 = HCF × LCM.  Hence we have
Observe that the bracketed number  (756/126)  is equal to  6  which is the numerator of the reduced fraction.  So we do not even need to make that calculation.  Just take the reduced numerator  6  and multiply that with the original denominator  1386  to get the LCM.  You may try a similar trick with  1386  and  126.  You end up multiplying  756  by  11,  which gives the same answer.

Further Exploration
Try the above with different pairs of numbers.  How can you extend this to find the HCF and LCM of three or more numbers?

(*) OK, just kidding.  In Singapore, we sort of follow British English, but we are flexible.

Suitable Levels
Lower Secondary Mathematics (Sec 1 ~ grade 7)
* other syllabuses that involve factors, HCF (GCD) and LCM
* any interested learner

## Monday, January 18, 2016

### [OlymLSec_20160118PPPC] A Square Proof by Contradiction

Question

Explanation
If  a + b = 11,  then  2ab = (a + b (a² + b²) = 121 100 = 21.  But  2ab  is an even number, whereas  21  is odd.  This is a contradiction.  So (B) is impossible.  ©

Remarks
Short and sweet isn’t it?  This uses the square-of-sum identity   (a + b= a² + 2ab + b².  I used the tactic of assuming the answer is correct  [H08]  and showing that this leads to something nonsensical [H05].  So the original assumption must be wrong.  This is called “proof by contradiction” or reductio ad absurdum (in Latin).
By the way, the correct answer option is (E) from the Pythagorean Triplet   8² + 6² = 10²  with  {a, b} = {8, 6}.  The question seems to be taken from some Kangaroo mathematics competition.

H05. Work backwards
H08. Make suppositions
H09. Restate the problem in another way
H13* Use Equation / write a Mathematical Sentence

Suitable Levels
Lower Secondary Mathematics competition
GCE ‘O’ Level “Elementary” Mathematics (challenge)
* other syllabuses that involve whole numbers and Pythagorean triplets
* any precocious or independent learner who loves a challenge

## Friday, January 15, 2016

### [AM_20160115SEAV] Simultaneous equations? Absolutely!

Problem / Question

Introduction
This question was from a Facebook group not dedicated to “Singapore math”.  The thing is, mathematics is really an international experience, especially with modern social media.
Simultaneous equations can pose a challenge to students, but this one is absolutely more challenging, because of the absolute value.

The Absolute Value

Strategy
The problem is: we do not know whether each of  x  and  y  is negative or otherwise.  That potentially raises complications.  There seems to be  4  cases to check.  However, by making assumptions  [H08]  separately and checking for contradictions [H07], we can narrow down the possibilities.  [H10]  Finally, we can simplify the problem to a regular pair of simultaneous linear equations   [H10]  and solve it by the method of elimination.  [H10, H11]

Solution

H04. Look for pattern(s)        [deciding what to eliminate]
H07. Use guess and check     [is x negative?  is  y negative?]
H08. Make suppositions        [is x negative?  is  y negative?]
H10. Simplify the problem
H11. Solve part of the problem
H13* Use Equation / write a Mathematical Sentence

Suitable Levels
GCE ‘O’ Level Additional Mathematics, “IP Mathematics” (challenge)
* other syllabuses that involve simultaneous equations and absolute values
* any precocious or independent learner who wants his/her mind tickled

## Saturday, January 9, 2016

### [OlymLSec_20160109CBDR] Derangement of Cars in a Roundabout

Problem / Question

Introduction
Some enthusiastic student posted this “deranged” question on Facebook.  It is taken from the Singapore Math Kangaroo Contest.  The question involves derangements, which does not appear in any syllabus in Singapore before university level.
If the problem involved a circular permutation, the correct answer would have been  (5 – 1)! = 4 × 3 × 2 × 1 = 24.  But this is not a circular permutation, because the roads are distinguished by their different positions.  If one rotates the roundabout, this would be considered a different pattern.  If it were a permutation, the answer would have been  5! = 5 × 4 × 3 × 2 × 1 = 120.  However, it is not a permutation, because the cars cannot go back to their original road in the opposite direction.  That is what the Kangaroo phrase “drives less than one round” mean.
A derangement is a some rearrangement in which things are not allowed to go back to their original position.  If one has not learned the formula for derangements, how could this be solved?

Strategy
We can consider simpler cases [H10] of the problem and build up the answer from there.  We may also split the problem into two cases.  [H11]

Solution

Let  !n  denote the number of derangements if there are  n  cars and  n  road branches in the roundabout.  [This is called the subfactorial of  n.]

For  n = 1,  !n = 0  because the car has no way but to go back on the original road.

For  n = 2,  !n = 1  because the only way is for the two cars to swap.

For  n = 3:  Car  #1  has two choices: either road  #2  or road  #3.  But it cannot end up swapping roads with any car, else the remaining car would have to go back to its original road, which is not allowed.  So it is either  #1 ® #2 ® #3 ® #1  or  #1 ® #3 ® #2 ® #1.  Therefore  !3 = 2.

For  n = 4:  Car  #1  has 3 choices.  For each of these choices, either it ends up (1) swapping roads with the other car, or  (2) it does not swap.
Case (1):  If it swaps with the other car, then the remaining  2  cars will have  !2 = 1 way.
Case (2):  If car  #1  goes to some road but not swapping roads with it,  then the other  3  cars will have  !3 = 2  ways to choose roads different from their original roads.
Therefore  !4 = 3 × (!3 + !2) = 3 × (2 + 1) = 9.

For  n = 5:  Car  #1  has 4 choices.
Case (1):  It swaps with the other car.  The remaining  3  cars will have  !3 = 2 ways.
Case (2):  It does not swap with any car.  The other  4  cars will have  !4 = 9  ways.
Therefore  !5 = 4 × (!4 + !3) = 4 × (9 + 2) = 44.

Ans:  (B)  44.

Remarks

H02. Use a diagram / model   (can be used for small number cases)
H03. Make a systematic list   (can be used for small number cases)
H04. Look for pattern(s)
H05. Work backwards
H09. Restate the problem in another way
H10. Simplify the problem
H11. Solve part of the problem

Suitable Levels
Lower Secondary Mathematics Competition
* University / College Combinatorics
* other syllabuses that involve derangements
* any precocious or interested learner who is interested

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