## Friday, 16 January 2015

### Class Discussion: When the condition changes...

1. Kevin, elaborate how you obtain the answer.

2. The multiplication between the biggest and smallest even root integer (or whatever that's called) must be the same with the multiplication of the middle 2 root integers, which is 12 and use that number to do this: 2^12 and we get 4096!

3. PS i just found that out today, Saturday 17 Jan 2015.

4. Originally, I did guess and check with calculator

5. @Kevin
I've difficulty trying to figure out what you try to explain (your post at 12:11).
Please elaborate with some numbers to illustrate.

1. 4096 group 3

1. Zi Bo, elaborate how you obtain the answer.

2. I obtained the answer by doing my index notations.i started of with 1^6,but it gives me 1 and the number is greater than 1.So,i cancelled 1 off my list.Next was 2^6,which gives me 64,but 64 cannot be square rooted.So,i cancelled 64 off my list.Next was 3^6,which gives me 729,but 729 also cannot be square rooted,so i cancelled 729 off my list.Next was 4^6,which gives me 4096,which can be square rooted,cube rooted,forth rooted and sixth rooted,and it is the smallest possible number.That is how i got my and.The steps may seem long,but i spent only 2 minutes doing it.

3. @Zi Bo
You are solving the problem by systematic elimination. Good.
On the other hand, now we have learnt LCM and HCF, do you think you would be able to apply your knowledge and skills learnt to solve this problem?

2. Group 2 cheat.they change their answer

1. Zi Bo,
perhaps you could think of a question to ask, that would help find out if Group 2 is able to justify the answer?

1. Vikhe, elaborate how you obtain the answer.