Bunuel wrote:
stne wrote:
Question does require some clarification for me.
Lets just focus on B
k is not divisible by any odd integer greater than 1
I can be sure of only one thing here, that is K is even, or it can be raised to the power of 2.
What is the definite value of K cannot be obtained by statement 2, under any circumstances, I think all of us agree.
e.g, K could be 2, 4, 8,16,32 etc
Now question is
Is k equal to 2^r for some positive integer r?
If K=2 and r=2,then is 2 = 2^2 answer is No 2 \neq 4 { there is no restriction on the value of k other than it has no odd factors and r has no restriction other than it is an integer, so k=2 and r=2 are both valid}
If K=4 and r=3,then is 4 = 2^3 answer is No 4 \neq 8 { there is no restriction on the value of k other than it has no odd factors and r has no restriction other than it is an integer, so k=4 and r=3 are both valid}
If K=4 and r=2,then is 4 = 2^2 Answer is Yes 4 = 4 { there is no restriction on the value of k other than it has no odd factors r has no restriction other than it is an integer, so k =4 and r=2 are both valid}
So we can see depending upon r ,2^rchanges .
so we can get both a Yes and a No depending upon K and r, can we not?
We do not have a definite K and a definite r, from 2
If question could have stated does K have only 2 as prime factors then of course solution could be more justified. I think some of us do agree that solution is debatable.
Lets just focus on B
k is not divisible by any odd integer greater than 1
I can be sure of only one thing here, that is K is even, or it can be raised to the power of 2.
What is the definite value of K cannot be obtained by statement 2, under any circumstances, I think all of us agree.
e.g, K could be 2, 4, 8,16,32 etc
Now question is
Is k equal to 2^r for some positive integer r?
If K=2 and r=2,then is 2 = 2^2 answer is No 2 \neq 4 { there is no restriction on the value of k other than it has no odd factors and r has no restriction other than it is an integer, so k=2 and r=2 are both valid}
If K=4 and r=3,then is 4 = 2^3 answer is No 4 \neq 8 { there is no restriction on the value of k other than it has no odd factors and r has no restriction other than it is an integer, so k=4 and r=3 are both valid}
If K=4 and r=2,then is 4 = 2^2 Answer is Yes 4 = 4 { there is no restriction on the value of k other than it has no odd factors r has no restriction other than it is an integer, so k =4 and r=2 are both valid}
So we can see depending upon r ,2^rchanges .
so we can get both a Yes and a No depending upon K and r, can we not?
We do not have a definite K and a definite r, from 2
If question could have stated does K have only 2 as prime factors then of course solution could be more justified. I think some of us do agree that solution is debatable.
I think you misunderstood the question.
The question basically asks: can k be written as some power of 2. (2) implies that k is some power of 2, thus it's sufficient.
Bunuel question says is K = 2^r
if K = 4 and r=3 then how is 4 = 2^3
The stress is on " equal to "