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Source: — Data Sufficiency |

by DanaJ » Fri Jan 23, 2009 2:24 pm
1 is not sufficient. We do not have any clues concerning k: it could be whatever, even a multiple of n = 20891....
2 is sufficient. k+6 and k+7 have only one common divisor: 1. Say x divides k+6. Dividing k+7 by x will give you a remainder of 1, no matter what the value of x is.

Answer: B
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