How to love [Singapore] math, without liking it?

In reply to "how to motivate a contemporary teenager to love Maths? Three key words," posted in the "Singapore Math NOW" LinkedIn Group, my reply then was:

Make the most disliked subject in school inaccessible, expensive and illegal.

Here are three less-irreverent practical suggestions that have proved to be successful to a certain degree, in depicting mathematics, not as a terminal subject once it becomes optional, but as a subject worthy of study, which can help raise one's socio-economic status.

1. Reward students with money. 

A brochure to motivate citizens to
upgrade themselves mathematically.  

In Singapore, mature students (who are generally working adults who are given a second or third chance to get a formal education qualification) get a few hundred bucks from self-help groups, or from some workforce development agency, if they attain a certain level of mathematical proficiency.

In other words, Singapore tax payers reward them with cash if they do well in math, on top of their heavily subsidized part-time education. The country wants them to succeed even if they'd previously made a mess of their "mathematical life."

2. Brand the subject or topic.

In the early eighties, leveraging on some pedagogical insights from research done in mathematics education from countries, such as China, Russia, Japan, and the United States, Singapore had succeeded in extending the power and application of the strategy, "Draw a diagram" or "Draw a model," to solve a whole set of mathematics problems, which traditionally were mostly solved by analytical or algebraic means.

As a result, local math curriculum developers, headed by Dr. Kho and his team, had subliminally or subconsciously branded the "draw a diagram" strategy, by re-christening it, the "model method," or the "bar method," as it's commonly known locally.

Today, the Singapore model method is an integral part of the Singapore math curriculum, as it's being powerfully used as a visualization strategy to solve a number of challenging word problems at the elementary school level, which would normally require a formal algebraic knowledge to solve them.

3. Value math highly in society.

Numeracy is not an option.

Provide opportunities for those with math degrees to hold positions of power and influence in the country, who may act as role models. 

For instance, many key personnel in the Singapore civil service (e.g., members of Parliament, army heads, and school principals) and in the private sector (e.g., CEOs, pastors, and TV actors) have a strong mathematical background, or are quantitatively (and financially) literate. It's probably no surprise that Singapore has the world's highest percent of millionaires per capita, probably because of a significant number of its citizens and residents being blessed with an above-average financial quotient (FQ).  

On 6/9/11, I jokingly tweeted a new definition of "Singapore math":

@MathPlus: Singapore math: A term to describe the ambivalent governance of Singapore by a Prime Minister and a President, who are both math majors.

And, on 15/2/12, Republic of Math tweeted the following:

@republicofmath: Singapore president Tony Tan has applied math PhD. Prime Minister Lee Hsien Loong has math degree http://t.co/0UHDpIuE via @JohnAllenPaulos

It's probably not a coincidence to attribute good governance and economic success in Singapore to applied quantitative literacy, partly because at the top management in both government and the private sector, we've in place key personnel who can think rationally and logically—not easily swayed by race or religion politics, and who would make unpopular decisions for the benefit of the country even at the risk of not being elected in the next election.

Math as a Servant 

Just like we can love our enemies, without liking them (if we yield ourselves to the guidance of the Holy Spirit), so too can we learn to appreciate and love the most-disliked subject in school, without having quite a liking for it. Like money, we need to make math our servant, not our master—we can make math work for us, without being enslaved by it. 

© Yan Kow Cheong, July 18, 2013.

The Chickens-and-Cows Problem

At the end of an earlier post entitled "The Chickens-and-Rabbits Problem," I tickled readers whether the following question could be solved using the Singapore model method, or the Sakamoto method.

Mr. Yan has almost twice as many chickens as cows. 

The total number of legs and heads is 184. How many cows are there?

On his "Johnny and Mary Do Maths" blog, Chris Patterson followed up with an algebraic solution to the word problem in a post entitled Visualisation versus Algebra?. He asked me whether I have a visual proof to the question, particularly one using the Sakamoto method.

I'm not sure whether I've succeeded in answering Chris's question, but here are three quick-and-dirty non-algebraic attempts to the chickens-and-cows problem.


Method 1 (Singapore model method)

From the model,

11 units = 184 + 3 ▌= 16 × 11 + (8 + 3 ▌)

If ▌= 1, then 11 units = 16 × 11 + 11

          1 unit = 17

2 units – 1 = 2 × 17 – 1 = 33

A quick check shows that for other integral values of ▌other than 1, 11 units cannot be an integer.

Hence, Mr. Yan has 33 chickens and 17 cows.


As a student of the Singapore model method, still trying to learn how to make wise use of this visualization strategy, especially when it lends itself well to a particular word problem, I would like to hear from members of the Singapore mathematical brethren about their alternative models in solving this chickens-and-cows problem, assuming that we're discussing this question with a group of grade 5 or 6 students—with no knowledge of Diophantine equations or advanced algebraic techniques.


Method 2 (Sakamoto method)

1. Grasp the relation

Let ① represent the number of cows, and △ be a variable quantity that is much less than ①.

                                           Chickens            Cows

                                              ② – △                ①

Number of legs                 (② – △) × 2         ① × 4

                                            = ④ – 2△            = ④

                                   _________________________

                                                           184

2. Diagram

3. Number sentences

② – △ + ④ – 2△ + ① + ④ = 184
⑪ – 3△ = 184
⑪ = 184 + 3△ = 16 × 11 + (8 + 3△)

If △ = 1, then ⑪ = 16 × 11 + 11 = 17 × 11
                       ① = 17
② – △ = 2 × 17 – 1 = 33

So, there are 17 cows and 33 chickens.


Unless there's a more elegant Sakamoto solution, I find that it doesn't differ much conceptually from its algebraic cousin.


Method 3 (Using the "Make a supposition" strategy)

Suppose there were exactly twice as many chickens as cows.

Then each group of 2 chickens and 1 cow would have a total of [(2 + 2 × 2) + (1 + 1 × 4)] = 11 legs and heads.

Now, 184 = 11 × 16 + 8

16 groups of 2 chickens and 1 cow would have a total of 16 × 11 = 176 legs and heads.

How many chickens and cows have 8 legs and heads altogether?

Clearly, 1 chicken and 1 cow have 8 legs and heads altogether.

So, Mr. Yan has (16 × 1 + 1) = 17 cows and (16 × 2 + 1) = 33 chickens.


Conclusion

With Chris's algebraic solution, we're sharing four methods of solution to the chickens-and-cows problem. Which method do you prefer? Which one would you use with your students? Share yours with the rest of us.


© Yan Kow Cheong, April 24, 2013.

The Dogs-and-Ducks Problem

Recently, while reviewing some grade 5 olympiad questions for a local company, which specializes in conducting olympiad math programs in elementary schools, I came across the following nonroutine dogs-and-ducks question.



The number of ducks is 10 fewer than the number of dogs. 
The dogs have three times as many legs as the ducks. 
How many ducks and how many dogs are there?



When I realized that my mental calculated answer differs from the one given by the problem poser, I thought maybe my incorrect solution might serve as a "logic exercise" to tickle others' mathematical bones to point out the flawed reasoning. My faulty argument followed something along these lines:

Since there are 10 fewer ducks, and each duck has 4 legs, there are 10 × 4 = 40 fewer duck legs.



From the model drawing,



2 units = 40


1 unit = 20



Number of duck legs = 1 unit = 20


Number of ducks = 20 ÷ 2 = 10



Number of dog legs = 3 units = 3 × 20 = 60


Number of dogs = 60 ÷ 4 = 15



A quick check shows that the answers don't satisfy the given conditions in the question. Although the dogs have three times as many legs as the ducks, however, the number of ducks is only 5 fewer than the number of dogs, which is expected to be 10.




When the why is harder than the how



The lesson here is to identify the illogical step, although seasoned problem solvers would probably not commit this kind of error or blunder. 

In general, it's one thing not to make an error, and get the correct answer; it's another thing to be able to explain someone else's error.

At a deeper level, it's sometimes conceptually harder to explain someone's flawed reasoning to an incorrect answer than just know how to solve a problem. Think of mathematical fallacies and paradoxes, where in a number of cases the illegal step is anything but obvious.




A fabricated faux solution

Alternatively, if we argued that a dog has 2 more legs than a duck, we'd then have a model that looks like this:

In this case, students would easily realize that the answers are also incorrect, when they find out that the number of dogs isn't an integer.


Two quick-and-dirty solutions

Why don't you try solving the dogs-and-ducks question on your own first, before comparing your solution with the ones I've worked out? If yours is an alternative (or elegant) solution, the mathematical brethren couldn't wait to reading it!

Method 1

Since a dog has twice as many legs as a duck, and there are 3 times as many dog legs as duck legs, there are 3/2 times as many dogs as ducks.

A model depicting the above relationship may be drawn as follows:

From the model drawing,

1 unit = 10

Number of ducks = 2 units = 2 × 10 = 20

Number of dogs = 3 units = 3 × 10 = 30

Check: Ducks: 20 × 2 = 40 legs

Dogs: 30 × 4 = 120 legs = 3 × 40 legs

Method 2

Given: Number of dogs = Number of ducks + 10

Number of duck legs = 2 × Number of ducks

Number of dog legs = 4 × Number of dogs

Given: Number of dog legs = 3 × Number of duck legs

A quick-and-dirty model representing the above information may be drawn as follows:

From the model drawing,

3 × 2 units = 6 units = 4 units + 10 + 10 + 10 + 10 

2 units = 10 + 10 + 10 + 10

1 unit = 10 + 10 = 20

1 unit + 10 = 30

Therefore, there are 20 ducks and 30 dogs.

Method 3 (Sakamoto method)

Those of you who are versed with the three-step Sakamoto method in solving word problems, may proceed as follows:

1. Grasp the relation

Let ④ represent the number of duck legs.

Duck legs           Dog legs

④                     ⑫

__________________________

Number of ducks     Number of dogs

④ ÷ 2 = ②                 ⑫ ÷ 4 = ③

                                             – 10

__________________________

1              :              1

                             + 10

2. Diagram

3. Number sentences

③ – ② = ① = 10

② = 2 × 10 = 20 (ducks)

③ = 3 × 10 = 30 (dogs)

The number of ducks is 20, and the number of dogs is 30.

Two bonus problems

If you like being challenged by similar dogs-and-ducks or chickens-and-rabbits problems, may I direct you to solving two more questions at Yan's One Minute Math Blog?

Happy Mathematical Problem Solving!


© Yan Kow Cheong, March 3, 2013