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Is K an integer? (1) K=P/Q, where P and Q are consequtive

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Is K an integer? (1) K=P/Q, where P and Q are consequtive [#permalink] New post   15 May 2003, 05:09
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Is K an integer?

(1) K=P/Q, where P and Q are consequtive integers;
(2) K=N^N, where N is an integer

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Re: Is K an integer? (1) K=P/Q, where P and Q are consequtive [#permalink] New post   15 May 2003, 05:53
Some are still alive...

Yes, it is E. :band
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Re: Is K an integer? (1) K=P/Q, where P and Q are consequtive [#permalink] New post   15 May 2003, 05:58
Ans : E

1) if P and Q +ve, ans N
if P and Q -ve, ans Y

2) if N +ve, ans Y
if N -ve ans N

Combined. There is NO solution for P/Q=N^N
(so K cannot be an integer but there is also no solution)
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Re: Is K an integer? (1) K=P/Q, where P and Q are consequtive [#permalink] New post   15 May 2003, 06:02
tzolkin wrote:
Ans : E

1) if P and Q +ve, ans N
if P and Q -ve, ans Y

2) if N +ve, ans Y
if N -ve ans N

Combined. There is NO solution for P/Q=N^N
(so K cannot be an integer but there is also no solution)


The trick with (1) is not positive vs. negative, but 0/1 vs. 7/8
Also Q cannot be a zero -- a sufficient condition to say no.

(2) what if N=0?
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Re: Is K an integer? (1) K=P/Q, where P and Q are consequtive [#permalink] New post   15 May 2003, 06:13
for

1) say p=-2 and q=-1 then P/Q = 2 (this is why I mentioned + -ve)

2) 2^2 = 4
-2^(-2) = -1/4

Combined
0^0 = 1 and no P/Q will be 1

Any thoughts ?

Why can't q=0 ? -1,0,1,2 etc....
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Re: Is K an integer? (1) K=P/Q, where P and Q are consequtive [#permalink] New post   15 May 2003, 06:42
Be careful with 0^0. It is not always equal to a zero?
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Re: Is K an integer? (1) K=P/Q, where P and Q are consequtive [#permalink] New post   15 May 2003, 11:33
The discussion of 0^0 is very old. Euler argues for 0^0 = 1 since a^0 = 1 for a not equal to 0 . The controversy raged throughout the nineteenth century, but was mainly conducted in the pages of the lesser journals: Grunert's Archiv and Schlomilch's Zeitshrift. Consensus has recently been built around setting the value of 0^0 = 1.

If 0^0 = 0 then 0/1=0 then N^N = P/Q and ans would be C
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Re: Is K an integer? (1) K=P/Q, where P and Q are consequtive [#permalink] New post   15 May 2003, 23:51
tzolkin wrote:
The discussion of 0^0 is very old. Euler argues for 0^0 = 1 since a^0 = 1 for a not equal to 0 . The controversy raged throughout the nineteenth century, but was mainly conducted in the pages of the lesser journals: Grunert's Archiv and Schlomilch's Zeitshrift. Consensus has recently been built around setting the value of 0^0 = 1.

If 0^0 = 0 then 0/1=0 then N^N = P/Q and ans would be C



It is the first time I have heard that it is legal to say that 0^0=1. From my university course of math I remember that the case is uncertain. It can be 1, but not always, depending on a function.
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Re: Is K an integer? (1) K=P/Q, where P and Q are consequtive [#permalink] New post   02 Dec 2010, 13:54
My answer is A as P and Q are consecutive integers and 0 is not an integer that means we can have P=1,Q=2 P/Q not an integer .Sufficient

For statement 2 N^N where N is an integer we can have N -ve as well as +ve so not sufficient
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Re: Is K an integer? (1) K=P/Q, where P and Q are consequtive [#permalink] New post   02 Dec 2010, 14:05
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rite2deepti wrote:
My answer is A as P and Q are consecutive integers and 0 is not an integer that means we can have P=1,Q=2 P/Q not an integer .Sufficient

For statement 2 N^N where N is an integer we can have N -ve as well as +ve so not sufficient


First of all: zero is an integer. Moreover it's an even integer.

Next, the case of 0^0 is not tested on GMAT.

This is poorly written question (definitely not GMAT type), so I wouldn't worry about it.
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Re: Is K an integer? (1) K=P/Q, where P and Q are consequtive [#permalink] New post   28 Jul 2016, 19:45
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Re: Is K an integer? (1) K=P/Q, where P and Q are consequtive [#permalink]
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