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!

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.

@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?

I did it by using guess and check and albeit of common factors, since it all is the same no., yet a different dividing amt. And soon, as i was experimenting with the cube root, i got 4096

4096

ReplyDeleteGroup 2

Kevin, elaborate how you obtain the answer.

Delete> Reply to your own post.

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!

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

DeleteOriginally, I did guess and check with calculator

Delete@Kevin

DeleteI've difficulty trying to figure out what you try to explain (your post at 12:11).

Please elaborate with some numbers to illustrate.

4096 group 3

ReplyDeleteZi Bo, elaborate how you obtain the answer.

Delete> Reply to your own post.

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.

Delete@Zi Bo

DeleteYou 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?

Group 2 cheat.they change their answer

ReplyDeleteZi Bo,

Deleteperhaps you could think of a question to ask, that would help find out if Group 2 is able to justify the answer?

4096

ReplyDeleteGroup 4

Vikhe, elaborate how you obtain the answer.

Delete> Reply to your own post.

4096

ReplyDeleteGRP 4

Marcus, elaborate how you obtain the answer.

Delete> Reply to your own post.

I did it by using guess and check and albeit of common factors, since it all is the same no., yet a different dividing amt. And soon, as i was experimenting with the cube root, i got 4096

Delete@Marcus

DeleteLet's think again, are we able to find a systematic manner to do this?

(i.e. not using guess and check method)