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Re: If k is an integer, is 2^k + 3^k = m ? [#permalink]

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30 Aug 2013, 00:26

danzig wrote:

If k is an integer, is 2^k + 3^k = m ?

(1) 4^k + 9^k = m^2 - 12

(2) k = 1

I don't agree with the OA. If we use statements 1 and 2 together, we get this:

We replace the value of \(k\) in the original question, so the question now is:\(is k = 5 ?\)

Now we replace the value of \(k\) in statement (1), so:

\(m^2 = 25\)

So,\(m = +/- 25\)

There is not indication whether m is possitive.

IMO, the answer: E

I don't know what was the original OA but I am assuming it was C, here's how I think the answer should be C, please do correct if there is a flaw in my reasoning.

Given If k is an integer, is \(2^k + 3^k = m\)? Squaring both sides original equation now becomes \(2^{2k} +3^{2k}+2.6^k=m^2\)

statement 1:\(4^k + 9^k = m^2 - 12\)this can be written as \(2^{2k} +3^{2k}+12 =m^2\)..

Comparing this with original equation we see that this will be equal to original equation only if K=1 since we do not have the value of K, hence insufficient

statement 2 : K=1 , by itself it is insufficient as we do not know the value of M

1+ 2

K=1 then \(m^2\) = 25

this is also what we get from the original equation,doesn't matter what m is, \(m^2\) is 25 Please share your views
_________________

I don't agree with the OA. If we use statements 1 and 2 together, we get this:

We replace the value of \(k\) in the original question, so the question now is:\(is k = 5 ?\)

Now we replace the value of \(k\) in statement (1), so:

\(m^2 = 25\)

So,\(m = +/- 25\)

There is not indication whether m is possitive.

IMO, the answer: E

I don't know what was the original OA but I am assuming it was C, here's how I think the answer should be C, please do correct if there is a flaw in my reasoning.

Given If k is an integer, is \(2^k + 3^k = m\)? Squaring both sides original equation now becomes \(2^{2k} +3^{2k}+2.6^k=m^2\)

statement 1:\(4^k + 9^k = m^2 - 12\)this can be written as \(2^{2k} +3^{2k}+12 =m^2\)..

Comparing this with original equation we see that this will be equal to original equation only if K=1 since we do not have the value of K, hence insufficient

statement 2 : K=1 , by itself it is insufficient as we do not know the value of M

1+ 2

K=1 then \(m^2\) = 25

this is also what we get from the original equation,doesn't matter what m is, \(m^2\) is 25 Please share your views

It does matter. If m=-5, then the question is: does \(2^k + 3^k = -5\)? And you cannot square this.
_________________

Re: If k is an integer, is 2^k + 3^k = m ? [#permalink]

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30 Aug 2013, 06:00

Bunuel wrote:

stne wrote:

danzig wrote:

If k is an integer, is 2^k + 3^k = m ?

(1) 4^k + 9^k = m^2 - 12

(2) k = 1

I don't agree with the OA. If we use statements 1 and 2 together, we get this:

We replace the value of \(k\) in the original question, so the question now is:\(is k = 5 ?\)

Now we replace the value of \(k\) in statement (1), so:

\(m^2 = 25\)

So,\(m = +/- 25\)

There is not indication whether m is possitive.

IMO, the answer: E

I don't know what was the original OA but I am assuming it was C, here's how I think the answer should be C, please do correct if there is a flaw in my reasoning.

Given If k is an integer, is \(2^k + 3^k = m\)? Squaring both sides original equation now becomes \(2^{2k} +3^{2k}+2.6^k=m^2\)

statement 1:\(4^k + 9^k = m^2 - 12\)this can be written as \(2^{2k} +3^{2k}+12 =m^2\)..

Comparing this with original equation we see that this will be equal to original equation only if K=1 since we do not have the value of K, hence insufficient

statement 2 : K=1 , by itself it is insufficient as we do not know the value of M

1+ 2

K=1 then \(m^2\) = 25

this is also what we get from the original equation,doesn't matter what m is, \(m^2\) is 25 Please share your views

It does matter. If m=-5, then the question is: does \(2^k + 3^k = -5\)? And you cannot square this.

Well if you say so, there are many ways algebraic equations can be manipulated and I thought they could be squared .Thank you +1
_________________

Re: If k is an integer, is 2^k + 3^k = m ? [#permalink]

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25 Dec 2014, 06:40

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