Chap 5 Alegbra... Algebra in Calendar (Group Work) - Solutions

On 19 May, the 4 groups presented their solutions (as shown below).
We learnt that:
  • There are many ways to the same problem. Collectively, the 4 groups presented a variety of methods to find answers to the 4 parts.
  • In some instance, there could be more than one way of writing the answer.
  • It is also important for us to be able to articulate the thinking process behind solving the problem. Some of the solutions are not 'self-explanatory'; hence being able to communicate our ideas and thought process clearly is important so that others understand.
  • To generalise a rule, we have to see beyond the 'rectangle' (i.e. beyond what is given) - this is obvious when we attempted part (e). Some of the groups even tested their observations (drawn from numbers within the box) by 'shifting' the rectangle, to see if what the observation stands when another set of numbers is selected.

To all the groups: A pat on your shoulder :D

Solution by Group 4: Shamus, Jian Hui, Jing Jie, Joshua Loh, Sue Lun [5 pts]

Solution by Group 3: Zhi Qi, Sher Li, Soh Fan, Mayur, Calvin, Guang Jun [9 pts]

Solution by Group 2: Ziying, Cheng Ngee, Joshua Ma, Shawn, Ming En [9 pts]

Solution by Group 1: Grace, Harsh, Kai Chek, Bryan, Jonathan, Zhi Chao [6 pts]


~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Here is one way of presenting the solutions...

(a) One way to carry out the calculation quickly is...
With reference to the pairs of numbers circled:
(7 + 23) + (8 + 22) + (9 + 21) + (14 + 16) + 15
= 30 x 4 + 15
= 135

(b) The sum of the 9 numbers
Solution A:
Sum of the 9 numbers
= n + (n+1) + (n+2) + (n+7) + (n+8) + (n+9) + (n+14) + (n+15) + (n+16)
= 9n + 72
Solution B:
Notice the pattern in the 1st column: 7, 14 and 21?
Sum of the 9 numbers
= n + (n+1) + (n+2) + 2n + (2n+1) + (2n+2) + 3n + (3n+1) + (3n+2)
= 18n + 9

(c) Sum of the 9 numbers
= (m-8) + (m-7) + (m-6) + (m-1) + m + (m+1) + (m+6) + (m+7) + (m+8)
= 9m
(d)
Solution A:
9m = 9n + 72
Dividing both sides by 9, we get... m = n + 8
Solution B:
9m = 18n + 9
Dividing both sides by 9, we get... m = 2n + 1

(e) Other interesting properties within that rectangle of numbers include:
  • The sum of the 9 numbers is a multiple of 9
  • The sum of the 3 numbers on each diagonal is equal