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I think this is a poor-quality question and I don't agree with the explanation. (2) @@ is divisible by 9. can be 0 or 9 (note that zero is divisible by every integer except zero itself). Not sufficient. So it could be 12340 or 12349 but these are not divisible by 9. It should be 12348, because the sum of 1+2+3+4+8 is divisible by 9.

The statement talks of @, the units digit, to be div by 9 so it has to be 0 or 9.. why are you confusing it with 1234@ being div by 9..
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(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.

Can someone please explain (1) to me? How is it that 0! works if it meets the criteria that it is NOT divisible by 5? You can divide 0/5. So that means @ would have to equal 1, 2 3, or 4. Which means that combining 1& 2 wouldn't work and answer would be E.

(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.

Can someone please explain (1) to me? How is it that 0! works if it meets the criteria that it is NOT divisible by 5? You can divide 0/5. So that means @ would have to equal 1, 2 3, or 4. Which means that combining 1& 2 wouldn't work and answer would be E.

Getting lost on this point...thank you!

(1) says that @! (factorial of number @) is NOT divisible by 5. can be 0 because @! = 0! = 1 and 1 is NOT divisible by 5.
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(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.

Can someone please explain (1) to me? How is it that 0! works if it meets the criteria that it is NOT divisible by 5? You can divide 0/5. So that means @ would have to equal 1, 2 3, or 4. Which means that combining 1& 2 wouldn't work and answer would be E.

Getting lost on this point...thank you!

(1) says that @! (factorial of number @) is NOT divisible by 5. be 0 because @! = 0! = 1 and 1 is NOT divisible by 5.

OK brilliant thank you...Had no idea 0!=1, which changes everything, of course.

Bunnel The question stem tells us that @ is either (0 or 5) as N is divisible by 5 . I Think we can get a unique answer with statement A because 1 is not divisible by 5 whereas 5! is divisible by 5 . I cannot understand why you have checked for @ {1,2,3,4} as we have already narrowed our search down to @{0,5}.

Bunnel The question stem tells us that @ is either (0 or 5) as N is divisible by 5 . I Think we can get a unique answer with statement A because 1 is not divisible by 5 whereas 5! is divisible by 5 . I cannot understand why you have checked for @ {1,2,3,4} as we have already narrowed our search down to @{0,5}.

You should read question and solution more carefully. The question does NOT say that @ equals to 0 or 5. The question asks whether @ equals to 0 or 5. From (1) if @ is 0, then the answer to the question is YES but if @ is 1, 2, 3, or 4, then the answer to the question is NO.
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I think this is a poor-quality question and I don't agree with the explanation. wrong answer

The question is 100% correct. I suggest to read carefully the solution and the whole thread once more. If you still have questions you are welcome to post.
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