Mastering Mathematics Smartly
by Wee Wen Shih

Main points:

• In the algebra method, we represent a problem using variable(s) and equation(s). We look for given pieces of information to form equation(s) with the variable(s).
  
  
• In the model method, we represent a problem using bar diagrams (with 1 unit clearly indicated) and equation(s). We look for given pieces of information to draw bar diagrams and form equation(s). Sometimes, shading may be necessary to indicate other ideas.
  
  
• In the algebra method, it is important to define as little variables as possible so that solving the problem becomes easier. One way is to look at the given information that highlights the relationship between unknowns.
  
  
• In the model method for a problem that involves before- and after-scenarios, we need to decide which situation to start with, so that solving the problem becomes easier. One way is to look at the given information that highlights the relationship between unknowns to aid us in making comparisons between bar diagrams.
  
  
• The concept of LCM (Least Common Multiple) of two or more numbers is important for solving a problem with the model method.
  
  
• It is common for us to make a comparison between the variable (of the algebra method) and the unit (of the model method).
  
  
• Recommended book: Top Marks Handbook by Dr. Fong Ho Kheong.

Main points:

• In the unitary method, we find the value of one unit or we compare the value of one part.
  
  
• Suggested hints to selected questions in Ex. 1.1 & Ex. 1.2.
  
  
• Recommended books from NIE library: Problem Solving the Systematic Way (Part 1) & (Part 2) by Markandoo P. & Vanessa Mah. Call number: QA43.56 Mar.

Main points:

• In percent problems, we need to be mindful of using the correct base in calculations.
  
  
• Clarification on the first example of "Speed Problems" (TG 3, please take note): In the algebra method, we let x = time taken by the car to travel 7 units to meet the lorry. So the time taken for the car to complete the rest of the journey (i.e. 5 units) after meeting the lorry is (5 - x) hours. Time taken is directly proportional to the distance travelled, so we obtain x : 5 - x = 7 : 5.
  
  
• Clarification on the car-tanker problem: We note that the tanker takes (4 2/5)h plus another (30/50)h to complete the remaining 5/9 of the journey. So 5h correspond to 5/9 of the journey. Hence, the tanker takes 5/5 x 9 = 9h for the entire journey.
  
  
• Solutions to Activity 2.

Main points:

• Tasks for e-learning week: Try to complete Activity 2, Activity 5 and Tutorial 2.1. We'll go through the details when we meet for a make-up lesson next Friday (5 Sep, either 12 - 2pm or 2.30 - 4.30pm, venue: 7-01-TR64).
  
  
• The Hindu-Arabic (H-A) numeration system has 10 symbols or numerals, is based on groupings of size 10 and is positional. Numbers from the H-A system are in base 10 and we could reuse the properties of the H-A system for representations in other bases.
  
  
• The expanded form is a sum of the products of face values and place values.
  
  
• In the base b system, the symbols are 0, 1, ..., b - 1 and place values are expressed in powers of b (i.e. groupings of size b). It will be good to review Activity 4 on your own.
  
  
• All tutorials are cancelled in view of the Learning Festival. We'll have consultations at the same 1-hour slots instead. We'll meet at NIE canteen level 2 for that.
  
  
• Suggested hint to Exercise 1.3 Q5: There are 2 scenarios we need to consider. In the first scenario, P starts at A and Q starts at B and both travel towards each other and meet in between A and B, say M. Thus, we see that distance AM : distance MB = 3 : 2. In order to travel 3 parts of the distance, P takes 60 minutes. Thus, P will take 100 minutes to travel the distance AB by proportional reasoning.
  
  In the second scenario, P starts at A and Q starts at B. P is going towards B and Q is going beyond B. Since P is faster than Q, P will eventually catch up with Q somewhere further than B, say N. Since the ratio of speeds remain the same (i.e. 3 : 2), we could use it for distance AB : distance BN. To travel AB, we already found that it takes 100 minutes. Now, we should be able to find the total time to travel AN.
  
  We could use the similar idea above for Q7.
  
  Important lesson to be learnt: Regardless of the meeting place (in between or beyond), the ratio of distances will follow the ratio of speeds.
  
  
• Suggested hint to Exercise 1.3 Q6: There are 2 scenarios to be considered. In the first scenario, P starts at A and Q starts at B and both move towards each other, meeting in between A and B, say M. Speed of P : speed of Q is 5 : 4, so distance AM : distance MB is 5 : 4.
  
  In the second scenario, speed of P : speed of Q becomes 4 : 4.8 due to 20% decrease and 20% increase respectively. From the meeting point M, P moves towards B while Q is heading to A. When P reaches B, Q is 10 km away from A, say N. Now distance MB : distance MN = 4 : 4.8. Notice that distance MN (4.8 parts) is part of distance AM (5 parts) of scenario 1. So the remaining 0.2 parts is equivalent to a distance of 10 km.

Main points:

• Happy teacher's day to all!
  
  
• Attempt last cohort's Quiz 1 (page 1, 2).
  
  
• Make-up lesson 1 is on Friday (5 Sep, either 12 - 2pm or 2.30 - 4.30pm, venue: 7-01-TR64).
  
  
• A brief article that discusses various numeration systems.
  
  
• An article "Exploring Mayan Numerals" that looks at the Mayan numeration system.
  
  
• Egyptian numeration system:
  - Has 7 symbols.
  - Makes use of the additive property.
  
  
• Babylonian numeration system:
  - Has 2 symbols.
  - Uses base 60. Place values are 60^0, 60^1, 60^2, 60^3 and so on. To find face values, we use repeated division by place values, starting with the largest place value.
  - Has a placeholder symbol for positions where no face value is present.
  
  
• Roman numeration system:
  - Has 7 symbols.
  - Makes use of the additive and multiplicative properties.
  - Makes use of the subtractive property when a smaller symbol comes before a larger one, i.e. IV (5 - 1 = 4), IX (10 - 1 = 9), XL (50 - 10 = 40), XC (100 - 10 = 90), CD (500 - 100 = 400), CM (1000 - 100 = 900).
  - Makes use of a multiplier of 1000 to represent large numbers. A bar is drawn above the numeral, e.g. IV bar represents 4000.
  
  
• Mayan numeration system:
  - Has 3 symbols, of which zero is defined.
  - Uses base 20(x18). Place values are 20^0, 20^1, (20^1)x18, (20^2)x18, (20^3)x18 and so on. To find face values, we use repeated division by place values, starting with the largest place value.
  - Any number is represented in a top-down fashion, starting at the level with the largest place value.

Main points:

• Operations in different bases:
  - Base b to base 10: use expanded notation.
  - Base 10 to base b: by repeated division of b and collecting all the remainders.
  - Addition of numbers in base b: regrouping is necessary when perfoming addition column-wise.
  - Subtraction of numbers in base b: renaming of columns and reducing by 1 are necessary when there is not enough to subtract.
  - Multiplication of numbers in base b: use the usual long multiplication principle.
  - Division of numbers in base b: use the usual long division principle.
  
  
• Numerals in base 12 are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, T, E. Typically, questions focus on numbers represented between base 2 and base 12. 
  
  
• Make-up lesson on Friday (12 Sep) at TR64 (12 - 2pm or 2.30 - 4.30pm). Please download in advance from Blackboard the set of Topic 3 notes. The set of presentation slides we went through, for your reference.

Main points:

• Additional questions and solutions to chapter 2 exercises.
  
  
• 2007 Quiz 2 question paper (pages 1 & 2)  and solutions, for your revision.
  
  
• 2007 Quiz 3 question paper (pages 1 & 2) and solutions, for your revision.
  
  
• We plan to complete chapter 3 this week, before we go for a break :)
  - View lesson 2 slides.
  - Practice question on proof by contradiction.
  - View lesson 3 slides.
  
  
• Make-up lesson on Friday (12 - 2pm or 2.30 - 4.30pm at TR64). We'll start on chapter 4, so please download and print the following documents in advance:
  - Chapter 4 summary
  - Chapter 4 worked examples
  - Exercise 4 questions
  Note: The content of these resources originated from the set of notes available from Blackboard. Whole details have been packaged deliberately in a student-friendly, bite-size fashion to focus on important points. Where time permits, the learner is encouraged to browse the set of notes on his/her own.
  
  - It is important to understand the various meanings of a | b (i.e. a divides b):
  1. a is the divisor, b is the dividend.
  2. a is a factor of b.
  3. b is divisible by a.
  4. b is a multiple of a.
  
  - It is useful to remember Theorem 2, Corollary 3, Theorem 4 and the factor tree approach of obtaining the prime factorization of a number. When a number n has a divisor b, the possible remainders are 0, 1, 2, ..., b - 1.
  
  - You may attempt Q1, Q2, Q3 and Q4 (first part) of our Exercise already.
  
  
• Two past year exam papers (05/06 & 06/07) have been put up on the BlackBoard, for your exam preparation. 
  
  
• Have a good break during recess week and study hard for Quiz 2! Take a read at this interesting speech by Adrian Tan, author of the Teenage Textbook.

Main points:

• Answers to Activity 2 (12-hour clock):
  No need to fill in the shaded portion because modulo addition and modulo multiplication are commutative.
  
  Additive identity = 0
  
  Multiplicative identity = 1
  
  Additive inverse for 0 = 0
  Additive inverse for 1 = 11
  Additive inverse for 2 = 10
  Additive inverse for 3 = 9
  Additive inverse for 4 = 8
  Additive inverse for 5 = 7
  Additive inverse for 6 = 6
  Additive inverse for 7 = 5
  Additive inverse for 8 = 4
  Additive inverse for 9 = 3
  Additive inverse for 10 = 2
  Additive inverse for 11 = 1
  
  Multiplicative inverse for 1 = 1
  Multiplicative inverse for 5 = 5
  Multiplicative inverse for 7 = 7
  Multiplicative inverse for 11 = 11
  
  x = 11
  
  
• Additional practice for chapter 3.

Main points:

• Wishing all muslim students a very Happy Hari Raya Aidilfitri!
  
  
• 2007 Quiz 4 QP (pp. 1, 2).
  
  
• During our 1-hour lesson, we covered the following content:
  1. Theorem 9 and its application to test for primality.
  2. Result from Activity 1 and its application to find the number of divisors and list them.
  3. Prime factorization and Euclidean Algorithm (repeated use of Theorem 11) to find GCD.
  
  
• Web link on Sieve of Erathosthenes, for your reference.
  
  
• An online GCD calculator that shows you clearly all the steps in the Euclidean Algorithm.
  
  
• Optional: This resource is for the mathematically inclined learner who wishes to understand the proof (by contradiction) of Theorem 9 (please refer to the Lemma near the end of page 1).
  
  
• Solutions to selected questions from Exercise 3, for your reference.

Main points:

• We covered the following during the 2-hour lesson:
  1. Prime factorization method to find LCM of two numbers.
  
  2. Application of Theorem 12 to find LCM of two numbers.
  This method requires the presence of GCD found earlier.
  
  3. Divisibility tests and application of Theorem 13.
  Divisibility test for 2, 4, 8: 2, 4, 8 divide last digit, last 2 digits, last 3 digits respectively.
  Divisibility test for 3, 9: 3, 9 divide sum of digits respectively.
  Divisibility test for 5, 10: for 5, last digit is 0 or 5; for 10, last digit is 0.
  
  In worked example 21, we had to resort to the trial and error approach to find possible pairs of a and b.
  
  
• Plan for 1-hour lesson: discuss areas of difficulty on chapter 4.
  
  
• Hints and answers to our version of Exercise 4:
  Q1(a) True. Apply Theorem 2.
  
  (b) False. Apply Corollary 3.
  
  Q2 Yes. Follow the approach of worked example 6.
  
  Q3 Straightforward. Composite.
  
  Q4 Straightforward. 24 factors. Try listing them as a practice.
  
  Q5 We did this in class. Essentially, we list down the 2-digit factors of 924. Then we locate the pair whose units digit has a '3'.
  Answer: 980.
  
  Q6(a) Straightforward. GCD = 15, LCM = 566 610.
  
  (b) Ignore it, not in syllabus after checking with the course coordinator.
  
  Q7 We did this in class. Essentially, a = (2)(3) or a = (2)(3)(p) where p is a prime and p is neither 2 nor 3.
  Answer: Possible values of a are (2)(3), (2)(3)(5), (2)(3)(7), (2)(3)(11), (2)(3)(13), making a total of 5 possibilities.
  Try listing the possible values of a if it lies between 1 and 300 inclusive of both 1 and 300.
  
  Q8(a) Use the divisibility test for 9. Remainder = 8.
  
  (b) Use Theorem 13 and follow the approach of worked example 21. Let me start you off below.
  
  Given: 15 | 437a2b.
  15 = (3)(5) and GCD (3, 5) = 1.
  By Theorem 13, 3 | 437a2b and 5 | 437a2b.
  You continue from this point on...
  Answers: b = 0, a = 2, 5, 8; b = 5, a = 0, 3, 6, 9.
  
  (c) Follow the approach of worked example 21.
  Answer: 43032 (thanks to Fazilah for pointing out my error!).
  Approach: We apply divisibility test for 11. Then we use trial and error on the expression |3 + a + b|.
  
  (d) Use Theorem 13. Not possible. Let me start you off below.
  
  Let the six-digit number be n.
  Given: 4 | n.
  4 = (2)(2), so 2 | n.
  Sum of digits in n = 1 + 2 + 3 + 4 + 5 + 6 = 21.
  You continue from this point on...
  
  
• This is a summary of chapter 3, for your reference. Burn and drink ya :P
  
  
• An additional resource on modular arithmetic, or the clock problem in chapter 3 (optional reading if your time permits).
  
  
• No lessons for weeks 11 and 12, as I will be away for in-camp training. We have already done the necessary make-ups :)
  
  
• On week 13 when I return, we'll do a 2-hour review (on last year's exam paper).

Try these questions:

Exam paper 02/03 (on BlackBoard)
Q2, Q3, Q9 as revision for test 4
Q6, Q10 as consolidation for topic 3

Exam paper 03/04 (on BlackBoard)
Q3, Q8, Q9 as revision for test 4
Q2(b), Q7 as consolidation for topic 3

• A student (thanks Prema!) highlighted 2 questions of doubt from exercise 3, which I'd like to share hints with you all. Good luck for Test 3!
  
  
• Hints to selected questions in exam paper 06/07:
  Q1(a)
  No, for Mayan system. Bottom numeral is disallowed.
  
  Q1(b)
  Base 10: ask yourself how many zeros are needed for the resulting numeral to be greater than 99999.
  Apply the same principle (i.e. how many Babylonian placeholders, Mayan placeholders) for the other two numeration system.
  
  Q3(d)
  We are given four factors 2, 5, 9 (3^2) and 10 (2 x 5). This means that the locker number has the form (2^a)(3^b)(5^c). We are also given that there are 12 factors in all, so (a + 1)(b + 1)(c + 1) = 12. Apply the method of trial and error to find a, b, c.
  
  Q4(a)(ii)
  No. Consider 1 / {3 x sqrt(2) } and ask yourself if it belongs to A.
  
  Q5(a)
  Prove by contradiction. Use the fact that Q is closed under addition.
  
  Q5(b)
  Prove by contradiction. Use the fact that Q is closed under multiplication.
  
  Q5(c)(i)
  S is closed under *.
  Explanation: a * b = GCD(3^i, 3^j) = 3^{min (i, j)} (by definition of GCD) which still belongs to S.
  
  Q5(c)(ii)
  1 is the identity.
  Try on your own to check that a * 1 = 1 * a.
  
  Q8(a)
  abcde = a x 10^4 + b x 10^3 + c x 10^2 + 10d + e
  = (9990a + 990b + 90c) + (10a + 10b + 10c + 10d + e)
  You continue from this point onwards...
  
  Q8(b)(i)
  For any 4-digit number abcd, the rule that is given (upon simplification) is 12 | (4a + 4b + 10c + d).
  abcd = a x 10^3 + b x 10^2 + 10c + d
  = (996a + 96b) + (4a + 4b + 10c + d)
  You continue from this point onwards...
  
  Q8(b)(ii)
  Use the same approach as (i). The extended rule is 4a + 4b + 4c + 10d + e or 4(a + b + c + d + e) + 6d - 3e if we want to follow the format given in the question.
  
  
• Hints and answers to Exam 02/03 and Exam 03/04 questions on topic 4:
  
  Exam 02/03 Q2
  (a) GCD = 1, LCM = 3947049
  (b) 3^0, 3^1, 3^2, 3^3, 3^4, 3^5
  
  Exam 02/03 Q3
  (a)(i) Not divisible by 9.
  (ii) Divisible by 15. Apply Theorem 13.
  (b)(i) Number is 43032. Approach: Apply divisibility test by 8, and start with first digit = 0.
  (ii) Number is 34012. Approach: Apply divisibility test by 11, and start with first digit = 0.
  
  Exam 02/03 Q9
  (a) Start by expanding abcd in base 10.
  (b) Remainder = 1. Approach:
  Let the first number be m and consider its last 2 digits, i.e. 89.
  Remainder when 89 is divided by 4 is 1. So m = 4a + 1 for some positive integer a, by Theorem 4.
  Let the second number be n and consider its last 2 digits, i.e. 21.
  Remainder when 21 is divided by 4 is 1. So n = 4b + 1 for some positive integer b, by Theorem 4.
  Consider mn = (4a + 1)(4b + 1) = 16ab + 4a + 4b + 1 = 4(4ab + a + b) + 1, giving us a remainder of 1.
  
  (c)(i) True if n is not 0.
  (ii) True. To justify, check that 2 | (n^2 + n) when n = 2k and when n = 2k + 1, where k is any positive integer.
  
  Exam 03/04 Q3
  (a) Composite, since 11 | 143.
  (b) (2^3)(3^1)(5^1), number of factors = (4)(2)(2) = 16
  (c) GCD = 4, LCM = 22512
  (d) 18. Apply Theorem 12.
  
  Exam 03/04 Q8
  (a) Condition 1: 3 | (22 + x + y)
  Condition 2: 11 | |x + y - 2| (i.e. absolute value of x + y - 2)
  Approach: Apply Theorem 13.
  
  (i) x = 0, y = 2
  Approach: Start with x = 0, then use trial and error for y by making sure that both conditions above are satisfied.
  
  (ii) x = 2, y = 0
  Approach: Look at condition 2 and note that x + y - 2 can only be zero, so that condition 1 may be satisfied at the same time.
  
  (b) Not in syllabus.
  
  Exam 03/04 Q9
  Approach: Expand abcd (base 5) in base 10, i.e. (a x 5^3) + (b x 5^2) + (c x 5) + d. Note that 10 (base 5) = 5 (base 10). Continue on your own...
  Rule: d = 0
  By the rule, 200 and 2310 (both in base 5) are divisible by 10 (base 5).
  
  
• Hints and answers to Exam 02/03 and Exam 03/04 questions on topic 3:
  
  Exam 02/03 Q6
  (a)(i) -2/sqrt(7)
  (ii) sqrt(7)/2
  (b) 1263/990
  (c)(i) Identity = b. Hint: Observe the 2nd row and 2nd column.
  (ii) Inverse of a = c, inverse of b = b, inverse of c = a
  (iii) Yes. Hint: Are table entries symmetrical about the diagonal?
  
  Exam 02/03 Q10
  (a) Yes. Verify that (a * b) * c = a * (b * c) for any a, b, c in E.
  (b) identity element = 0.
  (c) inverse of 2 = 2.
  (d) Does not exist.
  
  Exam 03/04 Q2
  (a) 21531/9900
  (b)(i) 101: yes, no, yes
  -3.5: no, no, yes
  sqrt(15): no, yes, yes
  22/7: no, no, yes
  
  (ii) 101: -101, 1/101
  -3.5: 3.5, -2/7
  sqrt(15): -sqrt(15), 1/sqrt(15)
  22/7: -22/7, 7/22
  
  Exam 03/04 Q7
  (a)(i) Line 1: + is associative.
  Line 2: + is commutative.
  Line 3: + is associative.
  Line 4: 0 is additive identity, so b + -b = 0.
  Line 5: 0 is additive identity, so 0 + -a = -a.
  
  (ii) Hint: Move the first term in (i) to RHS.
  (b) Apply the hints.
  (c) True. Use these properties:
  1. Division is multiplication of multiplicative inverse.
  2. Subtraction is addition of additive inverse.
  3. Multiplication is distributive over addition.
  
  
• Hints and answers to most of the questions in exam 07/08:
  
  Q1(a)
  Egyptian: To obtain the next number, multiply 20 to the current number.
  
  Mayan: To obtain the next number, add 4 to the current number.
  
  Babylonian: To obtain the next number, subtract 10 from the current number.
  
  Roman: 1st -> 2nd, we add 10. 2nd -> 3rd, we add 20. 3rd -> 4th, we add 30. 4th -> 5th, we add 40. 5th -> 6th, we add 50.
  
  Base 7: 1st -> 2nd, we add 3. 2nd -> 3rd, we add 4. 3rd -> 4th, we add 5. 4th -> 5th, we add 6. 5th -> 6th, we add 10 (base 7).
  
  Thus, the general approach is to find a pattern in the sequence of numbers.
  
  Q1(b)
  Base 7: 7^6
  Approach: We want to find n such that 2(7^n) > 99999.
  
  Egyptian: No place value.
  
  Babylonian: 60^3
  Approach: We want to find n such that 2 + 2(60) + 2(60^2) + ... + 2(60^n) > 99999.
  
  Mayan: (20^3)(18)
  Approach: We want to find n such that 2(20^n)(18) > 99999.
  
  Q2(a)
  (ii) ******43 R0.
  Approach: 161506411 = 161506403 + 5 = ******42 x 5 + 5 = (******42 + 1) x 5 = ******43 x 5.
  
  (iii) ******43 R3.
  Approach: We just add 3 to RHS of (ii), i.e. ******43 x 5 + 3 (see above). The division algorithm gives us a remainder of 3.
  
  (iv) ******41 R2.
  Approach: 161506400 can be obtained by subtracting 3 from (i).
  So, ******42 x 5 -3 = ******41 x 5 + 5 - 3 = ******41 + 2. The division algorithm gives us a remainder of 2.
  
  Q2(b)
  Sum = 1650 (base 7)
  Approach: Use the hint and form the 13 pairs (i.e. 1 & 66, 2 & 65, ..., 16 & 51). Each pair gives 100 (base 7). 13 pairs is equivalent to 16 (base 7) pairs. Thus, the sum is 100 x 16 + 50.
  
  Q2(c)
  10 x 3 = 30 (base 7), so "3" ought to be at the position with place value 7.
  Use the same argument to explain the position of "2".
  
  Q3(b)
  (i) 80
  (ii) 100 (Do not include rest time in the calculation! Thanks to Ralleeah for pointing out the error.)
  (iii) 80 (Include rest time in the calculation!)
  (iv) 11:05
  (v) 208 1/3
  
  Q4(a)
  (i) GCD = 97, LCM = 3760787.
  (ii) 97 x 283 (check that both numbers are prime)
  
  Q4(b)
  47000000000000736
  Approach: Use Theorem 13 and the fact that 72 = 8x9.
  
  Q5(a)
  314468/99900
  
  Q5(b)
  (i) Yes, Yes, No (e.g. 9/12 is not in S).
  (ii) 0, No identity, No identity.
  (iii) -3, No inverse, No inverse.
  
  Q6(a)
  (i) The student has given a mathematical argument based on just 3 numerical cases. For a mathematical argument to be valid, he/she has to prove it rigorously for it to be applicable in general.
  
  (ii) Approach: Use the hint and the facts that Q is closed under multiplication and that sqrt(6) is irrational.
  
  Q6(b)
  0.1212... = 12/99 = 24/198
  0.1313... = 13/99 = 26/198
  Since a lies strictly in between the two values, let a = 25/198 = 0.12626...
  Other alternative answers are acceptable, e.g. 49/396 (i.e. 0.123737...).
  
  Q6(c)(i)
  First line: -1 is the additive inverse of 1.
  Second line: Multiplication is distributive over addition.
  Third line: 1 is the multiplicative identity. (Thanks to Khai Han for pointing out my error.)
  
  Q6(d)
  Correct.
  Use these properties:
  1. Division is multiplication of multiplicative inverse.
  2. Multiplication is associative.
  
  Q7(a)
  abcd = (a x 10^3) + (b x 10^2) + (c x 10) + d, continue on your own...
  25 | abcd if 25 | (10c + d).
  
  Q7(b)
  (2^1)(3^2)(5^6) and many other possibilities.
  Approach: Say the number is N. N has 42 factors. We note that 42 = (2)(3)(7) = (1 + 1)(2 + 1)(6 + 1). So N has 3 prime factors whose powers are 1, 2, 6. Choose the prime factors to be the first 3 prime numbers, i.e. 2, 3, 5.
  
  Q7(c)
  Approach: Find GCD of both sales figures.
  Answers: 727 sold on Sunday, 363 sold on Monday.
  
  Q7(d)
  First, we look at the correct set of 3 numbers: 1125, 2925, N.
  1125 = (3^2)(5^3) and 2925 = (3^2)(5^2)(13^1)
  
  Next, we look at the misread set of 3 numbers: 1725, 2925, N.
  1725 = (3^1)(5^2)(23^1)
  
  Since both have the same LCM, this value = (3^2)(5^3)(13^1)(23).
  Thus, N = (5^3)(23) = 2875.

• Provide students with more opportunities to present in class, as one has to do mathematics in order to learn mathematics. Presentations will enable students to gain confidence in speaking in front of others.
  
  
• Assign problems in advance so that students could prepare their solutions on transparencies.
  
  
• Compile a list of past year examination questions for each topic to expose students early to the types of problem they will face.
  
  
• Continue to encourage students to use this website as a learning platform. Encourage students to discuss problems on the site's forum.
  
  
• Teach topic 3 more effectively to strengthen students' understanding of abstract concepts (binary operation, closure, commutativity, associativity, identity and inverse).