Tag Archives: problem solving

The 12 Problems of CHRISTmaths

Vintage Christmas—Just like Baby Jesus two millennia ago! Vintage Christmas—Just like Baby Jesus two millennia ago!

Christmas is a golden and joyful opportunity for number enthusiasts and math geeks to sharpen their creative mathematical problem-solving skills.

Here are 12 CHRISTmaths cookies that may help you shake your brain a little bit in the midst of Christmas festivities.

Warning: Refrain from forwarding this post to relatives or friends living in countries, which are intolerant of Christmas and Christianity, such as Brunei, Saudi Arabia, and Somalia, as it’s haram for “infidels” to take part in any kind of Christmas celebrations. And I assume that includes reading any on-line materials deemed un-Islamic or un-Mohammedan, which might lead believers astray from the faith.

1. Unlucky Turkeys

Estimate the number of turkeys that make their way to the supermarkets every year.

2. A Xmas Candy

Mary wanted to buy a candy that costs 25 cents. A dated vending machine would take one-cent, five-cent, and ten-cent coins in any combination. How many different ways can she use the coins to pay for the candy?

Christmastize your code!Remember to scan your Christmas item!

3. The Dimensions of a  Cross

A square of side 25 cm has four of its corners cut off to form a cross. What is the perimeter of the cross?

4. The Number of Crossings

Two lines can cross one time, three lines three times, four lines six times, and five lines ten times. If there are 25 lines, what would be the maximum number of crossings be?

5. An Eco-Xmas

If all instances of the word “CHRISTMAS” were replaced with “XMAS,” how much ink and paper (or Xmas trees) could you save every year? How much money could be channelled back to feeding the poor and the hungry during the festive season?

XMaths Tech© T. Gauld’s You’re all just jealous of my jetpack (2013)

6. Number of Xmas Cards

In an age of Xmas e-cards and video cards, how many Christmas greetings cards are still being sent worldwide? How many trees are being saved every festive season?

7. Does Xmas! have 25 digits?

1! = 1, 2! = 1 × 2 = 2, 5! = 1 × 2 × 3 × 4 × 5 = 120—a 3-digit number, and 10! = 1 × 2 ×⋯× 10 = 3,628,800—a 7-digit number.

(a) Without a calculator, how would you verify whether the number 25! has precisely 25 digits or not.

(b) Which positive integers n (other than the trivial case n = 1) for which n! has exactly n digits?

GST with no thanks to Father XmasGST (or VAT) with no thanks to Father Xmas

8. Xmas Trees

Guesstimate how big a forest would 25 million Christmas trees occupy.

9. Folding papers

Fold a single piece of paper perfectly in half, from left to right. How many creases will there be after the 25th fold, when you continue folding so that all the rectangles are folded into two halves each time?

10. Pre-Xmas Tax

Imagine Singapore were to implement a pre-Christmas tax on all kinds of Christmas marketing before the first week of December. Estimate how many extra million dollars would the Income Tax department collect every festive season.

Folding a Santa Claus© Anonymous Folding a Santa Claus

11. A Xmas Quickie or Toughie

What is the sum of the last two digits of 1! + 2! + 3! +⋯+ 24! + 25!?

12. An Ever-Early Xmas

Show that as one celebrates more and more Christmases (or, as one gets older and wiser), Christmas seems to come earlier every year.

Xmas Möbius Strips Christmas Möbius Strips

References

Gould T. (2013). You’re all just jealous of my jetpack. New York: Drawn & Quarterly.

Yan, K.C. (2011). Christmaths: A creative problem solving math book. Singapore: MathPlus Publishing.

Zettwoch, D., Huizenga, J., May, T. & Weaver, R. (2013). Amazing facts… & beyond! with Leon Beyond. Minneapolis: Uncivilized Books.

A Xmas Bonus: 25 CHRISTmaths Toughies from Singapore ?? http://tinyurl.com/q9w3ne9

 

Selected Hints & Answers

2. 12 ways. Hint: Make an organized list.

3. 100 cm. 

4. 300 crossings.

5. About 30 million gallons of ink, 500 square miles of paper, and $15 trillion could be saved.

6. Hint.

7. (b) n = 22, 23, 24.

9. 225 – 1.

11. 4.

12. Hint: Why as one gets older, time appears to fly faster.

2012-12-21 23.13.28

© Yan Kow Cheong, December 25, 2015.

 

The Numerology about Mr. Lee Kuan Yew

In the aftermath of the death of Singapore’s founding father, Mr. Lee Kuan Yew (1923–2015), a number of numerological tidbits (or numerical curiosities, to put it mildly) floated on social media, which got a number of apparently self-professed innumerates pretty excited. Here are three such postings that I saw in my Facebook feed and on WhatsApp.

RIP: Lee Kuan Yew (1923–2015)

A numerological message that was circulated among WhatsApp users in Singapore (© Unknown)—A numerological message that was shared among Singapore WhatsApp users

The WhatsApp message gives the impression that it was the works of some “pseudo-mathematician,” but it could very well have been the digital footprints of a “mathematical crank” or an amateur-numerologist, who wanted to tickle mathophobics with such numerical coincidences.

Did Singapore’s numerologists (or pseudo-mathematicians) fail to point out some of the following numerological absurdities?

The digital root of Mr. Lee’s birth year is 1 + 9 + 2 + 3 = 15, which stands for the last two digits of the year he experienced his last heartbeat.

The pollution index for that week was in an unhealthy range, and the average PSI for the six-day mourning period was about 91.

Or, were there exactly 91 priests on vigil at an undisclosed Roman Catholic Church, who were interceding for Mr. Lee to ensure that his heavenly destination is 100% secured, although his manifold deeds to the nation inarguably exceeds the number of his political faux pas, especially vis-à-vis his political enemies or opponents?

Or, did 91 senior monks and nuns (or did I mistake them for disciples of Shintoism?) resort to “synchronized chanting” to ensure that the highest level of enlightenment be bestowed on the late Mr. Lee, who might be reincarnated as a future Buddha for his numerous selfish deeds towards his oft-ungrateful and unappreciative fellow citizens?

And did any police personnel verify whether there were 91,000 odd mourners in black attire on that Black Sunday, not to say, 91 VIPs or Heads of States who attended the eulogy, depending on one’s definition of a VIP?

 

The Numerology of the Old Guard

One Facebook numerological factoid that circulated in the first post-LKY week was the following:

IMG_0817-0.JPG Singapore’s political fathers who outlived the biblical three-scores-and-ten lifespan

At face value, these nonagenarians had their blessed lives prolonged up to “four scores and ten and one” years. Sounds like their good earthly or political deeds were good karma for their longetivity? Are they the recipients of the following success equation?

Sacrifice + Service + Incorruptibility + Risk  = Political Success + Longevity

Observe that simply taking the difference between the birth year and the death year of Mr. S Rajaratnam suggests that he died at the age of 91; however, if we look closely at the month dates (Feb. 25, 1915 – Feb. 22, 2006), he was still 90 years old, when he passed away. The same argument goes for Dr. Toh Chin Chye (Dec. 10, 1921 – Feb. 3, 2012), who wasn’t yet 91, when he died. So, always take the pseudoscience of numerology with a grain of salt. As with fengshui charlatans, a degree of skepticism towards numerologists of all sizes and shapes isn’t an option—wear your critical-thinking cap when meeting, or reading about, these paranormal folks!

 

Fortune via Misfortune—From 4D to 5C

(© Unknown) Punters used combinations of the digits related to Mr. Lee death date to lure Lady Luck.

To rational non-punters or non-gamblers, betting on someone’s death date, whether he or she was poor or rich on this side of eternity, seems like an extreme case of bad taste,  or simply showing zero respect for the deceased and their family members. However, in superstitious circles, that practice isn’t uncommon among mathematically challenged or superstitious punters, who think that bad luck paranormally translates into good omen, if they bet on the digits derived from the death date or age of a recently deceased person.

In fact, during the nation’s six-day mourning period for its founder, besides the long queues of those who wanted to pay their last respects to Mr. Lee at the Parliament House, another common sight islandwide were meters-long lines of 4D or TOTO punters, who wanted to cash in on the “lucky digits” to retire prematurely, hoping to lay hold of the traditional 5Cs (cash, car, condo, credit cardcountry club), coveted by hundreds of thousands of materialistic Singaporeans.

 

Number Theory over Numerology

Fengshui in the Gym(© BBC) Chinese numerology in the gym? Or, is it just a mild form of deification of a political figure?

Instead of promoting a numerological or pseudoscientific gospel based on Mr. Kuan Yew’s death date or age, which only helps to propagate superstition and pseudoscience, perhaps a “mathematically healthy” exercise would be to leverage on the D-day to teach our students and their parents some basic numerical properties—for example, conducting a recreational math session on “Number Theory 101” for secondary  1–4 (or grades 7–10) students might prove more meaningful or fruitful than dabbling in some numerological prestidigitation, or unhealthy divination.

A Search for Patterns

91  is the product of two primes: 91 = 7 × 13

91 = 1² + 2² + 3² + 4² + 5² + 6²

91 is also the sum of three squares: 1² + 3² + 9²

Are there other ways of writing the number 91 as a sum of squares?

91 = 33 + 43

 

Non-Numerological Questions to Promote Problem-Solving Skills

Let’s look at an “inauspicious number” of elementary- and middle-school (primary 5–secondary 4) math questions, which could help promote numeracy rather than numerology among students and teachers.

1. Sum of Integers

Show that the number 91 may be represented as the sum of consecutive whole numbers. In how many ways can this be done?

2. The Recurring Decimal

What fraction represents the recurring decimal 0.919191…?

3. Palindromic in Base n

For what base(s) will the decimal number 91 be a palindromic number (a number that reads the same when its digits are reversed)? For example, 91 = 101013.

4. The Billion Heartbeat

Does a 91-year-lifespan last less or more than a billion heartbeats?

5. Day of the Week

Mr. Lee Kuan Yew (September 16, 1923–March 23, 2015) died on a Monday. Using the 28-year cycle of the Gregorian calendar, which day of the week was he born?

6. One Equation, Two Variables

If x and y are integers, how many solutions does the equation x² – y² = 91 have?

7. Singapore’s New Orchid

A new orchid—Singapore’s national flower—had been named after Mr. Lee: Aranda Lee Kuan YewUsing the code A = x, B = x + 1, C = x + 2, …, , does there exist an integer x such that ARANDA sums up to 91? In other words, does there exist a numerological system such that  A + R + A + N + D + A = 91?

8. Singapore’s Coin Goes Octal

Singapore's The alleged involvement of Mr. Lee in Singapore’s “lucky” octagonal one-dollar coin

There is an apocryphal story that had circulated for many years linking Mr. Lee Kuan Yew with Singapore’s octagonal one-dollar coin. A high-ranking monk had apparently told Mr. Lee that Singapore’s fortune would continue to rise only if Singaporeans were to carry a bagua—the eight-sided fengshui symbol. That prediction allegedly prompted the Monetary Authority of Singapore to issue the octagonal shape of the nation’s one-dollar coin.

That rumor was later put to rest by no other than self-declared agnostic Mr. Lee himself in one of his books, Hard Truths. He remarked that he had zero faith in horoscopes, much less the pseudoscience of fengshui.

What is the sum of the interior angles of the Singapore’s eight-sided coin?

9. Show that the largest number k for which the decimal expansion of 2k does not contain the digit 1 is 91.

© Yan Kow Cheong, April 26, 2015.

Resurrection isn't an option in Singapore!Resurrection isn’t an option in Singapore!

Selected Answers/Hints

1. One example is 91 = 1 + 2 + 3 +⋯+ 13.
2. 91/99.
5. Mr. Lee was born on a Sunday.
6. Hint: Show that x² – y² = 91 has 8 integer solutions.
9. Hint: Use a computer to verify the result.

Number Patterns on Social Media

If you had been on Facebook for some time, it’s very likely that you would have come across a certain type of numbers puzzles streaming your feed. Clearly, they’re not quite the typical numerical puzzles that often appear in school math books; rather, they’re closer to the types of questions posed in aptitude or IQ tests. One such number puzzle is the following.

20140611-133825-49105422.jpg

Unlike in textbooks that often present these logic puzzles in an uninteresting way, by seeing these colorful number puzzles on Facebook or Pinterest, and being hinted that only a small percent of the problem solvers apparently managed to get the correct answer, this entices readers to give it a try to see how well they’ll fare vis-à-vis their oft-mathematically challenged Facebook friends. Here’s another such number puzzle.

20140611-150315-54195123.jpg

Arguably, these numbers-and-words puzzles have a certain charm to it, because they often require just simple logic to solving them, unlike similar brainteasers that may require some knowledge of elementary or middle-school math. What about the following Facebook numerical puzzle?

20140611-150011-54011255.jpg

 

What is 1 + 1?

What do you make of this one?

20140611-154155-56515670.jpg

If these on-line numbers puzzles indirectly help promote logic and number sense among math-anxious social-media addicts; and along the way, provide them with some fun and entertainment, let’s have more of them in cyberspace!

Let me leave you with half a dozen of these numerical puzzles, stolen from my Facebook feed. Note that these types of brain-unfriendly math or logic questions may have more than one mathematically or logically valid answer, depending on the rule or formula you use—they serve as numerical catalysts for promoting creative thinking in mathematics.

20140612-114955-42595222.jpg

20140612-115024-42624274.jpg

20140612-115055-42655633.jpg

20140612-115202-42722803.jpg

20140612-115228-42748693.jpg

20140612-115317-42797136.jpg

Happy Creative Problem Solving!

© Yan Kow Cheong, June 12, 2014.

A Grade 5 Bicycles-and-Tricycles Problem

20140220-124202.jpg

In an earlier post, I shared about the following chickens-and-rabbits problem.

There are 100 chickens and rabbits altogether. The chickens have 80 more legs than the rabbits. How many chickens and how many rabbits are there?

Other than using a guess-and-guess strategy and an algebraic method, both of which offering little pedagogical or creative insight, let me repeat below one of the two intuitive methods I discussed then.

Since the chickens have 80 more legs than the rabbits, this represents 80 ÷ 2 = 40 chickens.

Among the remaining (100 – 40) = 60 chickens and rabbits, the number of chicken legs must be equal to the number of rabbit legs.

Since a rabbit has twice as many legs as a chicken, the number of chickens must be twice the number of rabbits in order for the total number of legs to be equal.

20140220-180550.jpg

From the model drawing,

3 units = 100 − 40 = 60
1 unit = 60 ÷ 3 = 20

Number of rabbits = 1 unit = 20
Number of chickens = 2 units + 40 = 2 × 20 + 40 = 80

The Bicycles-and-Tricycles Problem

Again, if we decided to ban any trial-and-error or algebraic method, how would you apply the intuitive method discussed above to solve a similar word problem on bicycles and tricycles?

There are 60 bicycles and tricycles altogether. The bicycles have 35 more wheels than the tricycles. How many bicycles and tricycles are there?

Go ahead and give it a try. What do you discover? Do you make any headway? In solving the bicycles-and-tricycles question, I find that there are no fewer of half a dozen methods or strategies, which could be introduced to elementary school students, three of which lend themselves easily to the model, or bar, method, excluding the Sakamoto method.

© Yan Kow Cheong, March 4, 2014.

Some Advice for Singapore Mathletes

20140205-183957.jpg

Here are some pointers I would share with my students at the start of a secondary math olympiad programme. In Singapore, most mathletes attending these enrichment classes are usually selected by the form or math teacher, who tends to choose the best three math students from each class to form a small group of 15 to 20 participants. They would then graduate to represent the school after attending a six-, eight-, or ten-session training programme, depending on the mathematical needs and wants of the school.

• Take your time! Very few contestants can solve all given problems within the time limit. For instance, in the Singapore Mathematics Olympiad (SMO), both at the junior and senior levels, [unconfirmed] feedback based on different schools’ results hints to the fact that those who can win a medal hover around five percent.

Interestingly but disturbingly, an SMO mathlete who can get six or seven out of 35 questions correct may still win a bronze medal, revealing how unmoderated olympiad math papers had been in recent years, going by the abnormally high rate of failures among the participants. This is primarily due to the fact that few, if any, faculty members who set these competition papers, are familiar with what elementary and middle school teachers are covering in local schools.

• Try the “easier” questions first. The questions aren’t set in ascending order of difficulty. It’s not uncommon to see easier questions in the second half of the paper.

• Olympiad questions don’t “crack” immediately. Be patient. Try various approaches. Experiment with simple cases. Working backwards from the desired result in some cases is helpful.

• If you’re using a contests book, and you’re stuck, glance at the “Hints” section. Sometimes a problem requires an unusual idea or technique.

• Even if you can solve a problem, read the hints and solutions. The hints may contain some ideas or insights that didn’t occur in your solution, and they may discuss intuitive, strategic, or tactical approaches that can be used elsewhere.

Remember that modeled or elegant solutions often conceal the torturous or tedious process of investigation, false starts, inspiration and attention to detail that led to them. Be aware of the behind-the-scenes hours-long dirty mathematical work! When you read the modeled solutions, try to reconstruct the thinking that went into them. Ask yourself, “What were the key ideas?” “How can I apply these ideas further?”

• Go back to the original problem later, and see if you can solve it in a different way, or in a different context. When all else fails, remember the reliable old friend, the guess-and-check strategy (or heuristic, as it’s being arguably called in Singapore)—for instance, substituting the optional answers-numbers given in an MCQ into some given equation or expression may yield the answer sooner than later.

Meaningful or creative problem solving takes practice, with insightful or elegant solutions not being the norm. Don’t get discouraged if you don’t seem to make any headway at first. The key isn’t to give up; come back to the question after a day or a week. Stickability and perseverance are two long-time buddies for full-time problem solvers.

Happy problem solving!

20140205-184012.jpg

© Yan Kow Cheong, Feb. 7, 2014.

The Lighter Side of Singapore Math (Part 5)

Elementary Math from an Advanced Standpoint

A Grade 4 Question

Ken has 69 planks that are of standard size. He would need 5 such planks to make a bookshelf. What is the most number of bookshelves Ken can make?

Method 1

Let x be the number of bookshelves Ken can make.

5x ≤ 69
5x/5 ≤ 69/5
x ≤ 13 4/5

So the maximum whole number that satisfies the inequality x ≤ 13 4/5 is 13.

Hence the most number of bookshelves is 13.

 

Method 2

Using the floor function, the most number of bookshelves is ⎣69/5⎦ = 13.

 

A Grade Two Question

Verify that 2 × 2 = 4.

Basic hint: Use the FOIL method.
Intermediate hint: Use area.
Advanced hint: Use axioms (à la Whitehead and Russell)

Solution

Using the FOIL method

2 × 2
= (1 + 1) × (1 + 1)
= 1 × 1 + 1 × 1 + 1 × 1 + 1 × 1
= 1 + 1 + 1 + 1
= 4   QED

 

A Grade Four Question

A rectangular enclosure is 30 meters wide and 50 meters long. Calculate its area in square meters.

Solution

20131228-144507.jpg

 

The My Pals Are Here Math Series

The My Pals Are Here series has been rumored to have been edited and ghostwritten by a hundred odd editors and freelancers in the last decade.

Lack of mathematical rigor was initially targeted against Dr. Fong Ho Kheong and his two co-authors by American profs in the first or/and second editions —probably by those who were “ghost advisors or consultants” for Everyday Math.

 

Deconstructing the Singapore Model Method

1. It’s a problem-solving strategy—a subset of the “Draw a diagram” strategy.

2. It’s a hybrid of China’s line method and Russia’s box (or US’s bar) method.

3. It’s the “Draw a diagram” strategy, which has attained a brand status in mathematics education circles.

4. It’s a problem-solving method that isn’t recommended for visually challenged or impaired learners.

5. It allows questions traditionally set at higher grades (using algebra) to be posed at lower grades (using bars).

 

Painless Singapore Math

Perhaps that’s how we’d promote Singapore math to an often-mathophobic audience!

20131222-143047.jpg

 

A Singapore Ex-Minister’s Math Book

20131229-155500.jpg

Dr. Yeo had the handwriting below depicted in his recreational math book, regarding his two granddaughters, Rebecca and Kathryn.

20131229-155535.jpg

 

References

Yan, K. C. (2013). The lighter side of Singapore math (Part 4). December 2013. http://www.singaporemathplus.com/2013/12/the-lighter-side-of-singapore-math-part.html

Yeo, A. (2006). The pleasures of pi, e and other interesting numbers. Singapore: World Scientific.

© Yan Kow Cheong, December 30, 2013.

The Singapore Excess-and-Shortage Problem

In Singapore, in grades four and five, there is one type of word problems that seldom fail to appear in most local problem-solving math books and school test papers, but almost inexistent in local textbooks and workbooks. This is another proof that most Singapore math textbooks ill-prepare local students to tackle non-routine questions, which are often used to filter the nerd from the herd, or at least stream the “better students” into the A-band classes.

Here are two examples of these “excess-and-shortage word problems.”

Some oranges are to be shared among a group of children. If each child gets 3 oranges, there will be 2 oranges left. If each child gets 4 oranges, there will be a shortage of 2 oranges. How many children are there in the group?

A math book costs $9 and a science book costs $7. If Steve spends all his money in the science books, he still has $6 left. However, if he buys the same number of math books, he needs another $8 more.
(a) How many books is Steve buying?
(b) How much money does he have?

A Numerical Recipe

Depicted below is a page from a grade 3/4 olympiad math book. It seems that the author preferred to give a quick-and-easy numerical recipe to solving these types of excess-and-shortage problems—it’s probably more convenient and less time-consuming to do so than to give a didactic exposition how one could logically or intuitively solve these questions with insight.

A page from Terry Chew’s “Maths Olympiad” (2007).

Strictly speaking, it’s incorrect to categorize these questions under the main heading of “Excess-and-Shortage Problems,” because it’s not uncommon to have situations, when the conditions may involve two cases of shortage, or two instances of excess.

In other words, these incorrectly called “excess-and-shortage” questions are made up of three types:
・Both conditions lead to an excess.
・Both conditions lead to a lack or shortage.
・One condition leads to an excess, the other to a shortage.

One Problem, Three [Non-Algebraic] Methods of Solution

Let’s consider one of these excess-and-shortage word problems, looking at how it would normally be solved by elementary math students, who have no training in formal algebra.

Jerry bought some candies for his students. If he gave each student 3 candies, he would have 16 candies left. If he gave each student 5 candies, he would be short of 6 candies.
(a) How many students are there?
(b) How many candies did Jerry buy?

If the above question were posed as a grade 7 math problem in Singapore, most students would solve it by algebra. However, in lower grades, a model (or intuitive) method is often presented. A survey of Singapore math assessment titles and test papers reveals that there are no fewer than half a dozen problem-solving strategies currently being used by teachers, tutors, and parents. Let’s look at three common methods of solution.

Method 1

20131026-231338.jpg

Difference in the number of candies = 5 – 3 = 2

The 16 extra candies are distributed among 16 ÷ 2 = 8 students, and the needed 6 candies among another 6 ÷ 2 = 3 students.

Total number of students = 8 + 3 = 11

(a) There are 11 students.

(b) Number of candies = 3 × 11 + 16 = 49 or  5 × 11 –  6 = 49

Jerry bought 49 candies.

Method 2

Let 1 unit represent the number of students.

20131027-213637.jpg

Since the number of candies remains the same in both cases, we have

3 units + 16  = 5 units – 6

20131026-231524.jpg

From the model,
2 units = 16 + 6 = 22
1 unit = 22 ÷ 2 = 11
3 units + 16 = 3 × 11 + 16 = 49

(a) There are 11 students.
(b) Jerry bought 49 candies.

Method 2 is similar to the Sakamoto method. Do you see why?

Method 3

The difference in the number of candies is 5 – 3 = 2.

20131026-231534.jpg

The extra 16 candies and the needed 6 candies give a total of 16 + 6 = 22 candies, which are then distributed, so that all students each received 2 extra candies.

The number of students is 22 ÷ 2 = 11.

The number of candies is 11 × 3 + 16 = 49, or 11 × 5 – 6 = 49.

Similar, Yet Different

Feedback from teachers, tutors, and parents suggests that even above-average students are often confused and challenged by the variety of these so-called shortage-and-excess problems, not including word problems that are set at a contest level. This is one main reason why a formulaic recipe may often do more harm than good in instilling confidence in students’ mathematical problem-solving skills.

Here are two grade 4 examples with a twist:

When a carton of apples were packed into bags of 4, there would be 3 apples left over. When the same number of apples were packed into bags of 6, there would still be 3 apples left over. What could be the least number of apples in the carton? (15)

Rose had some money to buy some plastic files. If she bought 12 files, she would need $8 more. If she bought 9 files, she would be left with $5. How much money did Rose have? ($44)

Conclusion

Exposing students of mixed abilities to various types of these excess-and-shortage word problems, and to different methods of solution, will help them gain confidence in, and sharpen, their problem-solving skills. Moreover, promoting non-algebraic (or intuitive) methods also allows these non-routine questions to be set in lower grades, whereby a diagram, or a model drawing, often lends itself easily to the solution.

References

Chew, T. (2008). Maths olympiad: Unleash the maths olympian in you — Intermediate (Pr 4 & 5, 10 – 12 years old). Singapore: Singapore Asian Publications.

Chew, T. (2007). Maths olympiad: Unleash the maths olympian in you — Beginner (Pr 3 & 4, 9 – 10 years old). Singapore: Singapore Asian Publications.

Yan, K. C. (2011). Primary mathematics challenging word problems. Singapore: Marshall Cavendish Education.

© Singapore Math, October 27, 2013.

PMCWP4-2See Worked Example 2 on page 8; try questions 7-8 on page 12.

Hungry ghosts don’t do Singapore math

In Singapore, every year around this time, folks who believe in hungry ghosts celebrate the one-month-long “Hungry Ghost Festival” (also known as the “Seventh Month”). The Seventh Month is like an Asian equivalent of Halloween, extended to one month—just spookier.

If you think that these spiritual vagabonds encircling the island are mere fictions or imaginations of some superstitious or irrational local folks who have put their blind faith in them, you’re in for a shock. These evil spirits can drive the hell out of ghosts agnostics, including those who deny the existence of such spiritual beings.

Picture

Hell money superstitious [or innumerate] folks can buy for a few bucks to pacify the “hungry ghosts.”

During the fearful Seventh Month, devotees would put on hold major life decisions, be it about getting married, purchasing a house, or signing a business deal. If you belong to the rational type, there’s no better time in Singapore to tie the knot (albeit there’s no guarantee that all your guests would show up on your D-Day); in fact, you can get the best deal of the year if your wedding day also happens to fall on a Friday 13—an “unlucky date” in an “unlucky month.”

Problem solving in the Seventh Month

I have no statistical data of the number of math teachers, who are hardcore Seventh Month disciples, who would play it safe, by going on some “mathematical fast” or diet during this fearful “inaupicious month.” As for the rest of us, let’s not allow fear, irrationality, or superstition to paralyze us from indulging into some creative mathematical problem solving.

Let’s see how the following “ghost” word problem may be solved using the Stack Method, a commonly used problem-solving strategy, slowing gaining popularity among math educators outside Singapore (which has often proved to be as good as, if not better than, the bar method in a number of problem-situations).

During the annual one-month-long Hungry Ghost Festival, a devotee used 1/4 and $45 of the amount in his PayHell account to buy an e-book entitled That Place Called Hades. He then donated 1/3 and $3 of the remaining amount to an on-line mortuary, whose members help to intercede for long-lost wicked souls. In the end, his PayHell account showed that he only had $55 left. How much money did he have at first?

Try solving this, using the Singapore model, or bar, method, before peeking at the quick-and-dirty stack-method solutions below.

Picture

From the stack drawing,
2 units = 55 + 13 + 15 + 15 = 98
4 units = 2 × 98 = 196

He had $196 in his PayHell account at first.

Alternatively, we may represent the stack drawing as follows:

Picture

From the model drawing,
2 units = 15 + 15 + 13 + 55 = 98
4 units = 2 × 98 = 196

The devotee had $196 in his account at first.

Another way of solving the “ghost question” is depicted below.

Picture

From the stack drawing,
6u = 55 + 13 + 15 + 15 = 98
12u = 2 × 98 = 196

He had $196 in his PayHell account at first.

A prayerful exercise for the lost souls

Let me end with a “wicked problem” I initially included in Aha! Math, a recreational math title I wrote for elementary math students. My challenge to you is to solve this rate question, using the Singapore bar method; better still, what about using the stack method? Happy problem solving!

Picture

How would you use the model, or bar, method to solve this “wicked problem”?
Reference
Yan, K. C. (2006). Aha! math! Singapore: SNP Panpac Education. 
© Yan Kow Cheong, August 28, 2013.

Picture

A businessman won this “lucky” urn with a $488,888 bid at a recent Hungry Ghost Festival auction.

A Before-and-After Singapore Math Problem

Picture

A Singapore math primer for grades 4–6 students, teachers, and parents

In Model Drawing for Challenging Word Problems, one of the better Singapore math primers to have been written by a non-Singaporean author for an American audience in recent years, under “Whole Numbers,” Lorraine Walker exemplified the following before-and-after problem, as we commonly call it in Singapore.

Mary had served $117, but her sister Suzanne had saved only $36. After they both earned the same amount of money washing dishes one weekend, Mary noticed she had twice as much money as Suzanne. What was the combined total they earned by doing dishes?

The solution offered is as follows:

Picture

© 2010 Crystal Springs Books

The author shared that she did two things to make the model look much clearer:

• To add color in the “After” model;
• To slide the unit bars to the right.

This is fine if students have easy access to colored pens, and know which parts to shift, but in practice this may not always be too convenient or easy, especially if the question gets somewhat more complicated.Let me share a quick-and-dirty solution how most [elementary math] teachers and tutors in Singapore would most likely approach this before-and-after problem if they were in charge of a group of average or above-average grades 4–5 students.

Picture

From the model drawing,

1 unit = $117 – $36 = $81
1 unit – $36 = $81 – $36 = $45

2 × $45 = $90

They earned a total of $90 by doing dishes.

Analysis of the model method

Notice that the placement of the bars matters—whether a bar representing an unknown quantity is placed before or after another bar representing a known quantity.

In our model, had we placed the [shaded] bar representing the unknown unit on the right, it would have been harder to deduce the relationship straightaway; besides, no sliding or shifting is necessary. So, placing the bar correctly helps us to figure out the relationship between the unknown unit and the known quantities easier and faster.

In general, shading and dotting the bars are preferable to coloring and sliding them, especially when the problem gets harder, with more than two conditions being involved.

The Stack Method

This word problem also lends itself very well to the Stack Method. In fact, one can argue that it may even be a better method of solution than the bar model, especially among visually inclined below-average students.

Take a look at a quick-and-dirty stack solution below, which may look similar to the bar method, but conceptually they involve different thinking processes. To a novice, it may appear that the stack method is just the bar method being depicted vertically, but it’s not. Perhaps in this question, the contrast isn’t too obvious, but for harder problems, the stack method can be seen to be more advantageous, offering a more elegant solution than the traditional bar method.

Picture

From the stack model diagram, note that the difference $81(= $117 – $36) must stand for the extra unit belonging to Mary.

1 unit = $81
$36 + ▅ = $81
▅ = $81 – $36 = $45
2 ▅ = 2 × $45 = $90

So, they had a total of $90.

The Sakamoto Method

This before-and-after problem also lends itself pretty well to the Sakamoto method, if the students have already learned the topic on Ratio. Try it out!

Let me leave you with three practice questions I lifted up from a set of before-and-after grades 4–6 problems I plan to publish in a new title I’m currently working on, all of which encourage readers to apply both the bar and the stack methods (and the Sakamoto method, if they’re familiar with it) to solving them.

Practice

Use the model and the stack methods to solve these questions.

1. At first, Joseph had $900 and Ruth had $500. After buying the same watch, Joseph has now three times as much money as Ruth. How much did the watch cost?

2. Moses and Aaron went shopping with a total of $170. After Moses spent 3/7 of his money and Aaron spent $38, they had the same amount of money left. How much money had Aaron at first?

3. Paul and Ryan went on a holiday trip with a total of $280. After Paul had spent 4/7 of his money and Ryan had spent $52, the amount Paul had left was 1/4 of what Ryan had left. How much money did Ryan have at first?

Answers
1. $300 2. $86 3. $196

Reference
Walker, L. (2010). Model drawing for challenging word problems: Finding solutions the Singapore way. Peterborough, NH: Crystal Springs Books.

© Yan Kow Cheong, August 4, 2013.

Problem Solving Made Difficult

Picture

The US edition of a grade 5 Singapore math supplementary title.

Recently, while revising a grade 5 supplementary book I wrote for Marshall Cavendish, I saw that other than the answer, there was no solution or hint provided to the following question.

If Ann gave $2 to Beth, Beth would have twice as much as Ann.
If Beth gave $2 to Ann, they would have the same amount of money.
How much did each person have?

Most grade 7 Singapore math textbooks and assessment books would normally carry a few of these typical word problems, whereby students are expected to use an algebraic method to solve them. For instance, using algebra, students would form two linear equations in x and y, before solving them by the elimination, or substitution, method. A pretty standard application of solving a pair of simultaneous linear equations, by an analytic method.

However, it’s not uncommon to see these types of word problems appearing in lower-grade supplementary titles, whereby students could solve them, using the Singapore model, or bar, method; and the Sakamoto method. In other words, these grade 7 and 8 questions could be solved by grade 5 and 6 students, using a non-algebraic method.

Algebra versus Model Drawing

Conceptually speaking, I think a grade 6 or 7 student who can solve the above word problem, using a model drawing, appears to exhibit a higher level of mathematical maturity than one who simply uses two variables to represent the unknowns, before forming two simultaneous linear equations to solve them. Of course, because the numbers in this question are relatively small, it’s not surprising to catch a number of average students relying on the trial-and-error method to find the answer.

Try to solve the question, using both algebra and a model; then compare the two methods of solution. Which one do you think demands a deeper or higher level of reasoning or thinking skills?

Depicted below is a model drawing of the above grade 5 word problem.

Picture

From the model drawing,

1 unit = 2 + 2 + 2 + 2 = 8
1 unit + 2 = 10
1 unit + 6 = 14

Ann had $10.
Beth had $14.

Generalizing the Problem

A minor change in the question, by altering the “number of times” Beth would have as much money as Ann, reveals an interesting pattern: the model drawing remains unchanged, except for the varying number of units that represent the same quantity.Here are two modified versions of the original grade 5 question.

If Ann gave $2 to Beth, Beth would have three times as much as Ann.

If Beth gave $2 to Ann, they would have the same amount of money.
How much did each person have?

Answer: Ann–$6; Beth–$10.

If Ann gave $2 to Beth, Beth would have five times as much as Ann.
If Beth gave $2 to Ann, they would have the same amount of money.
How much did each person have?

Answer: Ann–$4; Beth–$8.

From Problem Solving to Problem Posing

The two modified questions could serve as good practice for students to become skilled in model drawing, and to help them deduce numerical relationships confidently from them. Besides, they provide a good opportunity to challenge students to pose similar questions, by altering the “number of times” Beth would have as much money as Ann. Which numerical values would work, and what ones wouldn’t, in order for the model drawing to make sense, or for the question to remain solvable?

Conclusion

Let me end, by tickling you with another grade 5 question, similar to the previous three word problems.

If Ann gave $2 to Beth, Beth would have three times as much as Ann.
If Beth gave $2 to Ann, they would have twice as much money as Beth.
How much did each person have?

Answer: Ann–$4.40; Beth–$5.20.

How do you still use the model method to solve this slightly modified ratio question? Test it on your better students or colleagues! It’s slightly harder, because any obvious result isn’t easily deduced from the model drawing, as compared to the ones posed earlier on. Besides, unlike the three previous word problems whose answers are integers, this last problem has a decimal answer—it just doesn’t lend itself well to the guess-and-check strategy.

Share with us how your students or colleagues fare on this last question. Remember: No algebra allowed!

© Yan Kow Cheong, July 12, 2013.