Category Archives: Sakamoto method

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


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.


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

3 units + 16  = 5 units – 6


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.


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)


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.


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.

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

The Chickens-and-Rabbits Problem

In Singapore, the chickens-and-rabbits question was in vogue in the late nineties, when the Ministry of Education then wanted teachers to formally teach problem-solving strategies (or heuristics, as we commonly call them here). Two common methods of solution favored by local teachers are “guess and check” (for younger students) and “make a supposition.” And in recent years, as Sakamoto math strategies gain currency in more local and regional schools, we’ve been blessed with no fewer than three other methods of solution to solve this type of problems.

A Grade 5 Contest Problem

In math contests and competitions, it’s not uncommon to witness some variations of the chickens-and-rabbits problem, which often pose much difficulty even to students, who are fluent in the Singapore model method. Let’s look at a grade 5 chickens-and-rabbits question, with a slight twist.

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?

Give it a try first, before comparing your solution(s) with the ones I’ve exemplified below.

Method 1

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 for both their total number of legs to be equal.


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

A check shows that the answers do satisfy the conditions of the question.

Method 2

The equations resulting from the models for Methods 1 and 2 are the same, but conceptually this method is slightly different from the previous one.

The bar representing the number of chickens must be half the length of the bar representing the number of chicken legs. The bar representing the number of rabbits must be one quarter the length of the bar representing the number of rabbit legs.


From the model drawing,

3 units = 100 – 40 = 60
1 unit = 60 ÷ 3 = 20
2 units + 40 = 2 × 20 + 40 = 80

Therefore, the number of rabbits is 20, and the number of chickens is 80.

Let me leave you with another fertile chickens-and-legs problem, which should challenge most grade 5 or 6 students, not to say, their teachers and parents.

Mr. Yan has almost twice as many chickens as cows. The total number of legs and heads is 184. How many cows are there?

Could you use the bar method, or the Sakamoto method, to solve it?

© Yan Kow Cheong, March 3, 2013.