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Re: D0112 [#permalink]
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05 May 2016, 08:51
amilbusthon wrote: I think this is a poorquality 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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Re D0112 [#permalink]
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27 Jul 2016, 14:09
I think this is a highquality question and I agree with explanation.



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Re: D0112 [#permalink]
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08 Aug 2016, 20:18
Hello, regardin the first condition, why @ can't be 6 or 7? both ! are not divisible by 5.
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Re: D0112 [#permalink]
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09 Aug 2016, 01:43



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Re: D0112 [#permalink]
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10 Feb 2017, 21:19
This s an excellent question, tests logic very well and if done right, saves quite a bit time.



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Re: D0112 [#permalink]
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06 Mar 2017, 21:16
(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!



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Re: D0112 [#permalink]
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07 Mar 2017, 02:33
bananasss wrote: (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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Re: D0112 [#permalink]
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07 Mar 2017, 08:59
Bunuel wrote: bananasss wrote: (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.



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Re: D0112 [#permalink]
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15 Jun 2017, 06:54
@ is divisible by 9
You have to be very careful in GMAT. It is such a small thing to overlook.



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Re: D0112 [#permalink]
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01 Jul 2017, 08:01
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}.



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Re: D0112 [#permalink]
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01 Jul 2017, 09:21
akshaypareek312 wrote: 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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Re D0112 [#permalink]
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02 Oct 2017, 21:01
I think this is a poorquality question and I don't agree with the explanation. wrong answer



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Re: D0112 [#permalink]
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02 Oct 2017, 21:04



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17 Oct 2017, 20:19
I think this is a highquality question and I agree with explanation.



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Re D0112 [#permalink]
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19 Nov 2017, 06:46
I think this is a highquality question and I agree with explanation.



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Re D0112 [#permalink]
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25 Nov 2017, 02:48
I think this is a highquality question and I agree with explanation.



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Re: D0112 [#permalink]
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19 May 2018, 12:46
Bunuel wrote: If \(N = 1234@\) and \(@\) represents the units digit of \(N\), is \(N\) a multiple of 5?
(1) \(@!\) is not divisible by 5
(2) \(@\) is divisible by 9 This is a great question.







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