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# ds:

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Senior Manager
Joined: 02 Feb 2004
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25 Mar 2005, 06:44
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don't get it!
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Joined: 18 Nov 2004
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25 Mar 2005, 08:37
"E"

state 1....b is even......sp 10^a + b......if b = 2, then 10^a + 2 will always be divisible by 3 (as the sum of all digits will be 3) so ans is YES....
if b = 6, then 10^a + 6....if a is 1.....16 is not divisible by 3...insuff

state 2....b can only be from 1<b<10.....again insuff

combine....b can be 2,6...insuff
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27 Mar 2005, 08:15
anyone else want to take a crack at this?
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27 Mar 2005, 08:25
I will go with 'E'

The question is asking us whether 1+b is divisble by 3
Statement 1: b is an even integer.
if b is 2 b+1 will be divisbile by 3
if b is 6 b+1 will not be divisbile by 3,
So Statement 1 is not sufficient.

Statement 2: b < 10
if b is 8 b+1 will be divisbile by 3
if b is 6 b+1 will not be divisbile by 3
So Statement 2 is not sufficient.

Putting both together still it is not sufficient.
27 Mar 2005, 08:25
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