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i think unless premeditated....it would be very difficult to get this right with in 2 minutes...it took me more than 2 minutes still i got it wrong because i think there are too many cases to check and too many exceptions.... is it spoiling the test....poll may give better view....with how many got it right... thanks

I would beg to differ on the official answer 'C'...as the Question states.

N = 1234@ and @ represents the units digit, is N a multiple of 5?

(1) @! is not divisible by 5 (2) @ is divisible by 9

(1)@! is not divisible by 5 so @ cannot be 0 or 5==> N is not a multiple of 5==> can answer the question (2)@ is divisible by 9 so @ is 0 or 9; if it is 0 ==> N is a multiple of 5; if @ is 9 N is not a multiple of 5==>Not sufficient!

I choose A Please give me your comments or feedvack on this one.

I would beg to differ on the official answer 'C'...as the Question states.

N = 1234@ and @ represents the units digit, is N a multiple of 5?

(1) @! is not divisible by 5 (2) @ is divisible by 9

(1)@! is not divisible by 5 so @ cannot be 0 or 5==> N is not a multiple of 5==> can answer the question (2)@ is divisible by 9 so @ is 0 or 9; if it is 0 ==> N is a multiple of 5; if @ is 9 N is not a multiple of 5==>Not sufficient!

I choose A Please give me your comments or feedvack on this one.

Red part is not correct.

1233@ to be divisible by 5 symbol @ should represent either 0 or 5. So the question asks whether @ equals to 0 or 5.

(1) @! is not divisible by 5 --> @ can be 0, 1, 2, 3, or 4 (note that \(0!=1\)). Not sufficient. (2) @ is divisible by 9 --> @ can be 0 or 9 (note that zero is divisible by every integer except zero itself). Not sufficient.

(1)+(2) Intersection of the values for @ from (1) and (2) is @=0. Sufficient.

How can 4 be a possibility of "@"? 4! results in 24 and the question clearly states that "@" is a "ones digit" so you cant plug in 24 as a "ones digit" so the option narrows down to 0,1,2, or 3. Doesn't it?
_________________

i am still trying to grasp why for statement 2 to be correct, @ has to be 9? When you plug in 9, N is not divisible by 9, only 8 is! Am I just missing something here? Are there any quick trick to divisibility that I am not aware of? Please explain.

N = 1234@ and @ represents the units digit, is N a multiple of 5?

(1) @! is not divisible by 5 (2) @ is divisible by 9

======= (1)@! is not divisible by 5 so @ is not 0 or 5==> N is not a multiple of 5==> can answer the question (2)@ is divisible by 9 so @ is 0 or 9; if it is 0 ==> N is a multiple of 5; if @ is 9 N is not a multiple of 5==>Not sufficiency! I choose A Please give me the feedback !

Hi- can anyone answer this? why is the answer not A?

N = 1234@ and @ represents the units digit, is N a multiple of 5?

(1) @! is not divisible by 5 (2) @ is divisible by 9

======= (1)@! is not divisible by 5 so @ is not 0 or 5==> N is not a multiple of 5==> can answer the question (2)@ is divisible by 9 so @ is 0 or 9; if it is 0 ==> N is a multiple of 5; if @ is 9 N is not a multiple of 5==>Not sufficiency! I choose A Please give me the feedback !

Hi- can anyone answer this? why is the answer not A?

The red part is not correct: 0!=1, so from (1) @ can be 0.