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WinWinMBA
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G is the common factor of K+7 and n, n and k are positive 28 May 2005, 15:32
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G is the common factor of K+7 and n, n and k are positive integrator, G=


1 n=9753
2 n=k+6



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G is the common factor of K+7 and n, n and k are positive 28 May 2005, 18:22
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B

G=1

provided G cannot be -1

also I assumed that K is in fact k (typo) and integrator = integer.
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G is the common factor of K+7 and n, n and k are positive 28 May 2005, 23:46
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Hi Sparky,

Can you please explain this further and your thought process ?

Thanks
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G is the common factor of K+7 and n, n and k are positive 29 May 2005, 03:35
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WinWinMBA
G is the common factor of K+7 and n, n and k are positive integrator, G=


1 n=9753
2 n=k+6
Show more


1) insufficient n=9753=3251*3, k+7 can be any number that is divisible by 3251 or 3
2) k+6 and k+7, so in this case say k+6 is divisible by some integer G then (k+6)/G = t (an integer), and (k+7)/G = (k+6)/G + 1/G = t + 1/G. 1/G is an integer if G = 1 (or -1).
This is sufficient if G is positive.
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G is the common factor of K+7 and n, n and k are positive 29 May 2005, 09:21
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sparky
WinWinMBA
G is the common factor of K+7 and n, n and k are positive integrator, G=


1 n=9753
2 n=k+6

1) insufficient n=9753=3251*3, k+7 can be any number that is divisible by 3251 or 3
2) k+6 and k+7, so in this case say k+6 is divisible by some integer G then (k+6)/G = t (an integer), and (k+7)/G = (k+6)/G + 1/G = t + 1/G. 1/G is an integer if G = 1 (or -1).
This is sufficient if G is positive.
Show more



Good one sparky.

Since G can be 1 or -1, I think you could have put E as ans. Any reason why you chose B ?
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G is the common factor of K+7 and n, n and k are positive 29 May 2005, 13:58
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ashkg
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WinWinMBA
G is the common factor of K+7 and n, n and k are positive integrator, G=


1 n=9753
2 n=k+6

1) insufficient n=9753=3251*3, k+7 can be any number that is divisible by 3251 or 3
2) k+6 and k+7, so in this case say k+6 is divisible by some integer G then (k+6)/G = t (an integer), and (k+7)/G = (k+6)/G + 1/G = t + 1/G. 1/G is an integer if G = 1 (or -1).
This is sufficient if G is positive.


Good one sparky.

Since G can be 1 or -1, I think you could have put E as ans. Any reason why you chose B ?
Show more


I like to think 'positive' :)

from OG 'x is a divisor (factor) of y provided that y=xn for some integer n'

so I guess -1 counts then
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G is the common factor of K+7 and n, n and k are positive 30 May 2005, 08:12
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sparky
ashkg
sparky
[quote="WinWinMBA"]G is the common factor of K+7 and n, n and k are positive integrator, G=


1 n=9753
2 n=k+6

1) insufficient n=9753=3251*3, k+7 can be any number that is divisible by 3251 or 3
2) k+6 and k+7, so in this case say k+6 is divisible by some integer G then (k+6)/G = t (an integer), and (k+7)/G = (k+6)/G + 1/G = t + 1/G. 1/G is an integer if G = 1 (or -1).
This is sufficient if G is positive.



Good one sparky.

Since G can be 1 or -1, I think you could have put E as ans. Any reason why you chose B ?

I like to think 'positive' :)

from OG 'x is a divisor (factor) of y provided that y=xn for some integer n'

so I guess -1 counts then[/quote]


That was a good one sparky.
Is it now B or E

HMTG.



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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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Thank you for understanding, and happy exploring!
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