Hints For Students

How can I become a better problem solver?


The only way to become a better problem solver is problem solving.

Keep track of solutions to problems as opposed to just getting answers. Every good problem has more than one solution, all producing the same correct answer. The solutions can be beautiful or ugly; they can be short or take 8 pages in fine print; they can be based on guessing or reasoning etc. Ask your friends or parents about how they would solve a particular problem. Find other solutions in the books or on the internet. Explore a variety of possible solutions for every problem, comparing them to each other.

The best way to understand something is to explain it to somebody else. Whenever you obtained a solution to a problem (either by yourself, or from somebody else), try reproduce the solution for a friend or family member, and to explain why it works.

Be organized. Have all the solutions carefully written down, step by step, or printed out, so that you can refer to them later.


Do not get upset or discouraged about this: it is absolutely normal. These problems are specially designed to challenge even the best math students. Don’t give up. Take you time. Try to generate ideas. Print out the Problem Solver’s Guide and use it. With time and practice, you will learn to see many ways to solve problems that initially seem to be insoluble!

Problem Solver's Guide


Read the problem carefully. Make sure you understand what is given, and what is required to find.


Draw a picture or diagram. A good drawing can help solving any kind of problem, not only a geometry one. For geometry-related problems, a good drawing is crucial.


Try to remember all the mathematical concepts mentioned in the problem (”Perimeter? I certainly have heard of it in Standard 6, but I forgot what it is...”). Normally, if a problem introduces some advanced concept, all the neccessary definitions are given. If you are not sure about something, consult a textbook or a mathematical dictionary (this recommendation is of little use when you are solving problems at a contest, but at least you can do it while practicing at home).


Try to recognize known concepts that are present in the problem, but are not explicitely mentioned. For example, if you are dealing with a line that connects two adjacent sides of a rectangle, you might notice that it forms a right triangle.


Remember rules that may be applicable in the case. For example, if you are dealing with a right triangle, the Pythagorean Theorem applies to it, and also, its area can be found as half the product of its legs.


Sort out the values and quantities that you do not know but need to find, from ones that you already know, and ones that you do not need at all.


Develop a strategy for solving the problem: think of the steps that you are going to take, about the result of each step, and about the way this step will help you to find the solution.

If you cannot see all the way to the solution at once, try to perform some of the steps, and see if they are helpful. Some possible steps:

(a) If there is a quantity that you do not know, but need badly, denote it with a variable and try to set up an equation.

(b) If the problem is asking for a property of some big number, try to look at smaller numbers and see if there is a pattern.

(c) If you have ever seen a problem that was (at least in part) similar to this one, try to remember what approach helped solving that problem. Can it be extended for solving this problem? (Keeping your notes organized and storing solutions rather than answers really helps, as you will see!)

(d) If nothing else helps, try the Guess and Check approach. In some cases, this approach will help you to get the answer. Do not forget to revise it later, so that you have a solution, too.


Once you have figured out the answer, compare it to the question of the problem to see if it makes sense. For example, if the problem asked you about the number of students in the classroom, and you got −3, 6.25, or 13500, you probably have made a mistake: check your solution.


If the answer looks reasonable, plug it into the problem to doublecheck that it works.

A Favourite Rich Problem

Most leading educators agree that the best way to teach mathematics is through problem solving. Some go so far as to say that it is the only way. The National Council of Teachers of Mathematics has named problem solving as the very first Process Standard. The following illustrates just how rich a good problem can be. Six arrows land on a target consisting of 3 concentric circular regions. Each arrow earns a score of 3 or 5 or 7 points. Which of the following total scores can be possible: 16, 19, 26, 34, 41, 44? (adapted from the December 1983 Olympiad)

One solution: With six arrows the least possible total score is 18 and the greatest is 42. Thus the scores 16 and 44 are both impossible. Since each score is odd, the sum of any pair is even and the sum of all six scores is even. Thus totals of 19 and 41 are impossible. Therefore only 26 and 34 remain. Both are possible. 26 can be achieved as follows: {3,3,3,3,7,7}, {3,3,3,5,5,7}, {3,3,5,5,5,5} and 34 as follows: {7,7,7,7,3,3}, {7,7,7,5,5,3}, {7,7,5,5,5,5}.

Comments: The original problem gave 31 as one of its choices. Replacing 31 with 34 made a very rich problem even richer for class discussions. In order to solve the problem properly, students must utilize the six separate thoughts below drawn from four important concepts. Arithmetic skills are practiced, but they are subordinate to mathematical thinking.

1. Don’t stop when you see how to score one of the given totals. After all, the question does not state that only one choice satisfies the conditions.

2. Eliminate all impossibilities. The Process of Elimination is as invaluable in mathematics as it is in daily life.

3. Realize that the sum of any two odd integers is even. This powerful concept simplifies some difficult-to-explain problems.

4. Realize that the sum of any number of even integers is even. This is obvious but often overlooked.

5. Look for a minimum total. Because 3 or 4 or 5 is absurd, there must be a specific lower limit to the scores.

6. Similarly, look for a maximum total. Because 300 or 400 or 500 is absurd, there must be a top limit, too.

Buried under the surface of this problem are four important concepts: completeness, parity, setting limits, and changing one’s point of view.

Completeness — thoughts 1 and 2 above depend upon the fundamental attitude that every answer must be complete. Youngsters should be taught to “give the answer, the whole answer, and nothing but the answer”. To answer “26” but not “34” is wrong.

Parity — thoughts 3 and 4 establish the usefulness of parity — the relationships between odd and even numbers. Students who realize whether a sum or product is odd or even have learned to look for the structure beneath the surface.

Setting limits — thoughts 5 and 6 express the concept that the set of possible answers to this question is really quite limited. Teach students to automatically determine the floor and ceiling (18 and 42) for the set. In algebra later on, this concept of domain for possible answers will be utilized extensively.

Changing one’s point of view — thoughts 5 and 6 should be seen as two faces of the same concept. If you see a floor, look also for a ceiling. Creativity can come from reversing a concept, a powerful technique.

Usually, most students use guess-and-check at first, at least until they start to understand the “soul” of the question. This has two advantages. Without realizing it, students drill themselves in basic addition facts, and the concept is implanted in their minds that math is not always immediate. This helps develop persistence.

Explanation Of Terms


(a) Prime Number
A prime number is defined as any number that can be divided, without a remainder, only by itself and by 1.

For example: 13 (which is prime) can only be divided evenly (no remainder) by 13 and by 1.

Other helpful hints:

  • Zero and 1 are not considered prime.
  • The only even prime number is 2. All other even numbers can be divided by 2. And remember, if it ends in ZERO it's also even.
  • No number greater than the prime number 5 ends in a 5. Because, if it ends in a 5 can be divided by 5.
  • Any number where the "sum" of the digits add up to a number that is divisible by "3" is not prime.
    Example: The number 1,269 is not prime because the sum of the digits (1 + 2 + 6 + 9 = 18) and since "18" is evenly devisable by 3, the number 1,269 is NOT a prime number.
  • Except for 0 and 1, a number is either a prime number or a composite number. A composite number is defined as any number, greater than 1, that is not prime.

(b) GCF (Greatest Common Factor)

The greatest common factor of two or more whole numbers is the largest whole number that divides evenly into each of the numbers.

List all of the factors of each number, then list the common factors and choose the largest one.

Find the GCF of 48 and 60.

The factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, and 48.

The factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60.

The common factors of 48 and 60 are 1, 2, 3, 4, 6, 12

Although the numbers in bold are all common factors of both 48 and 60, 12 is the greatest common factor.

(c) LCM (Least Common Multiple)
The least common multiple of two or more non-zero whole numbers is actually the smallest whole number that is divisible by each of the numbers.


Find the LCM of 4, 6 and 15.

Multiples of 4 are: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, ... (add 4 to each to get the next multiple).

Multiples of 6 are: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, ... (add 6 to each to get the next multiple).

Multiples of 15 are: 15, 30, 45, 60, 75, 90, ... (add 15 to each to get the next multiple).

The least common multiple of 4, 6 and 15 is the smallest number that appear in each list: 60


(a) Cryptarithms
Cryptarithms are a type of mathematical puzzle in which the digits are replaced by symbols (typically letters of the alphabet).

Cryptarithm Rules :

  • Each letter represents a unique digit.
  • Numbers must not start with a zero.
  • The solution is unique (unless otherwise stated).

9567 + 1085 = 10652

can be represented like this:

abcd + efgb = efcbh