Question

friends plse help

see the ques number 11
http://crpfpsrohini.files.wordpress.com/2010/11/viii-class-paper1.pdf

Answer 1

If 72K is a perfect cube, what is the value of K ?
That question?

Answer 2

yes

Answer 3

It would help if you factor 72 into primes... can you do that for me? :)

Answer 4

2*2*2*3*3

Answer 5

6

Answer 6

6?

Answer 7

Let's leave it in exponential form, shall we?
\[\large 2^3\cdot 3^2\]
Now, for this to be a perfect cube, all exponents must be multiples of 3... so, you have a
\[\large 2^3\] that's good, but you also have a \(\large 3^2\) which has an exponent which is not divisible by 3... so how do you "fix" that? :)

Answer 8

one more 3 has to be taken

Answer 9

Yup... so the value of K is...?
;)

Answer 10

3

Answer 11

There you go :)

Answer 12

okay @terenzreignz
thanks bro

Answer 13

Another reasoning, equally correct, and yielding the correct answer. :)

Answer 14

is this right way to solve it

Answer 15

@terenzreignz

Answer 16

Yes... it's basically what we did, said in another way :)

Answer 17

okay thanks again

Answer 18

:)