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If k is an integer greater than 1, is k equal to \(2^r\) for some positive integer r?

(1) k is divisible by \(2^6\)

(2) k is not divisible by any odd integer great than 1

Please explain your answer.

Thanks

I will go for (B)

(1) When k= 2^8, then, Yes. When k= 3* 2^8, then No. Insuff (2) Any integer not divisible by an odd integer greater than 1 could be represented in the form 2^r. Hence suff. I couldn't think of any example otherwise.
_________________

To find what you seek in the road of life, the best proverb of all is that which says: "Leave no stone unturned." -Edward Bulwer Lytton

If k is an integer greater than 1, is k equal to \(2^r\) for some positive integer r?

(1) k is divisible by \(2^6\)

(2) k is not divisible by any odd integer great than 1

Please explain your answer.

Thanks

I will go for (B)

(1) When k= 2^8, then, Yes. When k= 3* 2^8, then No. Insuff (2) Any integer not divisible by an odd integer greater than 1 could be represented in the form 2^r. Hence suff. I couldn't think of any example otherwise.

I didn't understand your analysis of statement 2. Would you please explain how? thanks

If k is an integer greater than 1, is k equal to \(2^r\) for some positive integer r?

(1) k is divisible by \(2^6\)

(2) k is not divisible by any odd integer great than 1

Please explain your answer.

Thanks

I will go for (B)

(1) When k= 2^8, then, Yes. When k= 3* 2^8, then No. Insuff (2) Any integer not divisible by an odd integer greater than 1 could be represented in the form 2^r. Hence suff. I couldn't think of any example otherwise.

I didn't understand your analysis of statement 2. Would you please explain how? thanks

=> It is easy for us to find an integer greater than 1 that is not divisible by an odd interger (>1) and represent it in the form of 2^r. For example- 2^8 is not divisible by any odd integer. =>In order to prove that (2) is insuff, we have to find an integer which is not divisble by an odd integer and that cannot be represented in the form of 2^r. There is no number greater than 1 which satisfies this condition. Any integer which cannot be divisible by an odd integer is an even integer which can be represented in the form of 2^r.

Not sure, if I confused you more
_________________

To find what you seek in the road of life, the best proverb of all is that which says: "Leave no stone unturned." -Edward Bulwer Lytton

If k is an integer greater than 1, is k equal to \(2^r\) for some positive integer r?

(1) k is divisible by \(2^6\)

(2) k is not divisible by any odd integer great than 1

Please explain your answer.

Thanks

k = some positive integer greater than 1

the question stem is asking is k of the form \(2^r\). When will it be of this form ? when it has only 2 as its prime factor.

(1) k is divisible by \(2^6\). This just means k has \(2^6\) as its factor, but it does not say this is the ONLY factor. There could be some odd factors as well !

(2) k is not divisible by any odd integer > 1. This means k has only even factors. When you have only even factors, it can be written in the \(2^x\) form.

I pick B.
_________________

"You have to find it. No one else can find it for you." - Bjorn Borg

(2) k is not divisible by any odd integer > 1. This means k has only even factors. When you have only even factors, it can be written in the \(2^x\) form.

I pick B.

Let's say k has 6 as a factor. which means k = 6L with out knowing L we can't write k in the form of 2 ^ r for example if L=3, K=18 Can 18 be written as a factor of 2 ^ r? NO. if k=8L and L =4 K= 2 ^ 5.

(2) k is not divisible by any odd integer > 1. This means k has only even factors. When you have only even factors, it can be written in the \(2^x\) form.

I pick B.

Let's say k has 6 as a factor. which means k = 6L with out knowing L we can't write k in the form of 2 ^ r for example if L=3, K=18 Can 18 be written as a factor of 2 ^ r? NO. if k=8L and L =4 K= 2 ^ 5.

Am I doing any thing wrong?

The way i understood the statement in the problem, if k has only even factor, it has only even prime factors.

k=6 means k has 3 as one of its factors.

I may be completely wrong.
_________________

"You have to find it. No one else can find it for you." - Bjorn Borg

(2) k is not divisible by any odd integer > 1. This means k has only even factors. When you have only even factors, it can be written in the \(2^x\) form.

I pick B.

Let's say k has 6 as a factor. which means k = 6L with out knowing L we can't write k in the form of 2 ^ r for example if L=3, K=18 Can 18 be written as a factor of 2 ^ r? NO. if k=8L and L =4 K= 2 ^ 5.

Am I doing any thing wrong?

The way i understood the statement in the problem, if k has only even factor, it has only even prime factors.

k=6 means k has 3 as one of its factors.

I may be completely wrong.

Ok. I get it. I got it wrong. K cannot have 6 as a factor because it will automatically make 3 a factor and that is not in line with (2) That makes only 2 the remaining factor.