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divisibility ! confused !thanks

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Source: — Data Sufficiency |
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by MadhurG » Wed Sep 10, 2008 12:37 am
If we divide 4321 by 3 - remainder 1 - to make it divisible by 3
values of K (less than 10) are 2,5 and 8.

Stmt 1 insufficient

If we divide 4321 by 7 - remainder 2 - to make it divisible by 7
value of K (less than 10) is just 5.

Stmt 2 sufficient.

Hence B
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