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Question No 89 in OG 11

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by captcha » Mon Dec 20, 2010 10:36 pm
(1) K^1 is not equal to 1^k for some numbers k
is K^(l+m) = (k^l)+(k^m)
--> "^" is clearly "-". [As for "+" & "x" 1+k=k+1; & 1xk=kx1;]

So, k-(l+m) = (k-l)+(k-m)
=> k-l-m = 2k - l -m
(for all k except k=0, which is only 1 number atleast 2 values of k required to say "for some numbers")

Ans to original problem is no. Hence, Statement 1 alone is sufficient.

(2) As above if ^ is -, the two sides of the equation are not equal. The answer to original question is No & statement 2 alone is sufficient to deduce this.

Therefore, both statements are alone sufficient to solve the problem.

IMO D.

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by fskilnik@GMATH » Mon Dec 20, 2010 11:04 pm
smalltown wrote:If ^ represents one of the operators +, -, and x, is k^(l+m) = (k^l) + (k^m) for all numbers of k, l, and m ?

(1) k^1 is not equal to 1^k for some numbers k
(2) ^ represents subtraction.

But I think statement '2' alone cannot answer the question as the question clearly asks 'is k^(l+m) = (k^l) + (k^m) for all numbers of k, l, and m' ? As per statement 2, if k is zero, then answer is Yes. If k is not equal to zero, then answer is No. So I think answer should not be either 'B' or 'D' .

Could you please explain where I made mistake ? How to solve this kind of problems ?
What should we assume if question says 'for some numbers' ?
Hi smalltown!

When you are asked if a sentence is true for all numbers, you DECIDE that if:
you are able to prove that the sentence IS true for ALL numbers
OR
if you are able to prove that the sentence is NOT true for A PARTICULAR case!

In other words, sttm (2) is sufficient to answer the question asked because you can easily find ONE EXAMPLE of k, l and m such that k^(l+m) = (k^l) + (k^m) is NOT true when ^is the subtraction operation.

I hope things are clear now.

Regards,
Fabio.

P.S.: in this scenario, a sttm would NOT decide if you had the following situation: ^ could be at least 2 different operations, in the first one you could prove that it would answer in the affirmative, that is, the sentence would be true for all numbers k, l and m, but in the second option for ^ the sentence would be answered in the negative, that is, you would be able to find an example of k, l and m such that the sentence would not be true for that particular ^ operation.

P.S.2: please note that sttm (1) is sufficient to answer the question asked because from this sttm we know that the operation ^MUST BE (between the choices offered, for sure) the subtraction, because if it were the addition or multiplication, you would have k^1 = 1^k for all values of k!! From the fact that we know that ^is subtraction, we know that sttm (1) decides the question asked, as explained above.
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by Woozler » Wed Dec 22, 2010 11:50 am
They're asking you whether k^(l+m) = (k^l) + (k^m) is true for ALL numbers, and you, it seems, are trying to pick particular numbers (for statement 2) that may work (if one of them is zero). THIS IS NOT WHAT THE QUESTION IS ASKING YOU. They're asking you if it's true for ALL numbers, not "some".

Anyway, the only way that k^(l+m) = (k^l) + (k^m) can work for all k, l, m is if ^ is multiplication.

(1) clearly states that k^1 is NOT 1^k in all instances, which is a clear violation of the commutativity rule (a*b = b*a) that we all learned in middle school. Thus, ^ is not multiplication and (1) is sufficient.

(2) very clearly states that ^ is not multiplication, and is also sufficient.

Therefore, (D)
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by fskilnik@GMATH » Wed Dec 22, 2010 12:41 pm
Woozler wrote:Anyway, the only way that k^(l+m) = (k^l) + (k^m) can work for all k, l, m is if ^ is multiplication.
Important: Woozler´s statement is true if ^ is to be chosen among the operations offered in this question stem.
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