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If N=1234@, is N a multiple of 5?
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Updated on: 06 Jul 2017, 09:35
Question Stats:
28% (01:41) correct 72% (01:32) wrong based on 206 sessions
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If N=1234@, is N a multiple of 5? (1) @! is not divisible by 5 (2) @ is divisble by 9 The OA is C, but I can't understand why! My reasoning is as follows:
This question can be rephrased as: "is N divisible by 5"? For this to be the case, N must end in 5 or 0.
(1) tells us that 0<@<4, since the factorial of any integer greater than 4 is divisible by 5. Thus, N is divisible by 5 if @=0 but is not divisible by 5 when @=1, 2, or 4. INSUFFICIENT.
(2) If @ is divisibly by 9, @=9 since no singledigit, nonzero, positive integer is divisible by any number less than itself. If @=9=units digit of N, N cannot be divisible by 5. SUFFICIENT.
Hence, my answer is B. Anyone disagree? Where am I going wrong? OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/ifn1234an ... 02671.html
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Originally posted by carriedinterest on 04 Feb 2010, 08:05.
Last edited by Bunuel on 06 Jul 2017, 09:35, edited 2 times in total.
Renamed the topic, edited the question and added the OA.



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Re: If N=1234@, is N a multiple of 5?
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04 Feb 2010, 08:30
I think I see my mistake. From (2), we know that @ must be 9 OR 0, since (technically) 0/9=0.
Taken together, we know that:
0<@<4
and
@ = 0 or 9.
Thus, @ must equal 0. If @=0, N is divisible by 5. TOGETHER SUFFICIENT.
Tricky question  I would never have thought to consider 0/9 as a possiblity under timed conditions. I guess that's the benefit of practice!



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Re: If N=1234@, is N a multiple of 5?
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04 Feb 2010, 08:58
I also think that B is right.
But if 1234@ has to be divisible by 9 then @ has to be 8 (and not 9)
12348/9 = 1372 (complete division)



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Re: If N=1234@, is N a multiple of 5?
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04 Feb 2010, 15:26
carriedinterest wrote: I think I see my mistake. From (2), we know that @ must be 9 OR 0, since (technically) 0/9=0.
Taken together, we know that:
0<@<4
and
@ = 0 or 9.
Thus, @ must equal 0. If @=0, N is divisible by 5. TOGETHER SUFFICIENT.
Tricky question  I would never have thought to consider 0/9 as a possiblity under timed conditions. I guess that's the benefit of practice! I think the answer should be D, Lets consider option 1, its says @ ! is not divisible by 5 so you already mentioned 0<=@=<4 but here don't you think if @ is 0 then 0! is also divisible by 5 ( 0/5 =0) and hence from option 1 we know that @ = 1,2,3 or 4 but not 0 and hence we know that if N is not divisible by 5 ? Please let me know is my reasoning wrong here ? Cheers



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Re: If N=1234@, is N a multiple of 5?
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04 Feb 2010, 16:14
nitishmahajan wrote: carriedinterest wrote: I think I see my mistake. From (2), we know that @ must be 9 OR 0, since (technically) 0/9=0.
Taken together, we know that:
0<@<4
and
@ = 0 or 9.
Thus, @ must equal 0. If @=0, N is divisible by 5. TOGETHER SUFFICIENT.
Tricky question  I would never have thought to consider 0/9 as a possiblity under timed conditions. I guess that's the benefit of practice! I think the answer should be D, Lets consider option 1, its says @ ! is not divisible by 5 so you already mentioned 0<=@=<4 but here don't you think if @ is 0 then 0! is also divisible by 5 ( 0/5 =0) and hence from option 1 we know that @ = 1,2,3 or 4 but not 0 and hence we know that if N is not divisible by 5 ? Please let me know is my reasoning wrong here ? Cheers @! is not divisible by 5, means that @ can be 0, 1, 2, 3, or 4. What you should know is that \(0!=1\), which implies that @ can be 0 as well.
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Re: If N=1234@, is N a multiple of 5?
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04 Feb 2010, 16:22
Bunuel wrote: nitishmahajan wrote: carriedinterest wrote: I think I see my mistake. From (2), we know that @ must be 9 OR 0, since (technically) 0/9=0.
Taken together, we know that:
0<@<4
and
@ = 0 or 9.
Thus, @ must equal 0. If @=0, N is divisible by 5. TOGETHER SUFFICIENT.
Tricky question  I would never have thought to consider 0/9 as a possiblity under timed conditions. I guess that's the benefit of practice! I think the answer should be D, Lets consider option 1, its says @ ! is not divisible by 5 so you already mentioned 0<=@=<4 but here don't you think if @ is 0 then 0! is also divisible by 5 ( 0/5 =0) and hence from option 1 we know that @ = 1,2,3 or 4 but not 0 and hence we know that if N is not divisible by 5 ? Please let me know is my reasoning wrong here ? Cheers @! is not divisible by 5, means that @ can be 0, 1, 2, 3, or 4. What you should know is that \(0!=1\), which implies that @ can be 0 as well. Thanks Bunuel for pointing out my mistake ..! Cheers



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Re: If N=1234@, is N a multiple of 5?
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05 Feb 2010, 05:02
piyatiwari wrote: I also think that B is right.
But if 1234@ has to be divisible by 9 then @ has to be 8 (and not 9)
12348/9 = 1372 (complete division) I don't agree that 1234@ must be divisible by 9. From statement (2), we know that @ must be divisible by 9, which means that @ must equal 9 or 0 (since no other singledigit integer between 0 and 9 is divisible by 9). The question is whether 0/9 is a legitimate answer here if it is, we know from (1) and (2) together than @ must be 0, and hence that N is divisible by 5. If it isn't, we know that @ must be 9 and hence N cannot be divisible by 9. As the OA is C, I'm inclined toward the view that @ must be 0 and that the statements are TOGETHER SUFFICIENT.



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Re: If N=1234@, is N a multiple of 5?
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09 Feb 2010, 08:00
I'm sure that GMAT Club Diagnostic uses the same rules to structure its questions as does GMAC. Based on what I have read about GMAC rules in structuring Data Sufficiency questions Statement 1 can’t contradict Statement 2 and vise versa.
Statement 1 indicates that @ could be equal to 0, 1, 2, 3, or 4. Therefore, it is not sufficient along. Statement 2 indicates that @ can be either 0 or 9. Combining 2 statements together one can conclude that @=0
So C is the correct answer.
@ cannot be 9 as it would contradict Statement 1 by implying that 9! is not divisible by 5, which is not true.



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Re: If N=1234@, is N a multiple of 5?
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13 Feb 2010, 12:15
carriedinterest wrote: From the GMAT Club Diagnostic:
If N=1234@, is N a multiple of 5?
(1) @! is not divisible by 5
(2) @ is divisble by 9 Ques: If 1234@ is divisble by 5? S1: @! is not divisible by 5. This means all integers = 0,1,2,3,4. But out of all... if @ = 0, 1234@ is divisible by 5... but if @ = 1,2,3,4... then 1234@ is not divisible. Hence Not Suff. S2: @ is divisble by 9. This means integers = 0, 9. Again if @ =0, 1234@ is divisible by 5.. but if @ = 9, 1234@ is not divisible by 5. Hence Not Suff. Combining S1 and S2, @ = 0..... which gives 1234@ is divisible by 5. Therefore answer is C.
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Re: If N=1234@, is N a multiple of 5?
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13 Oct 2015, 06:43
carriedinterest wrote: If N=1234@, is N a multiple of 5? (1) @! is not divisible by 5 (2) @ is divisble by 9 The OA is C, but I can't understand why! My reasoning is as follows:
This question can be rephrased as: "is N divisible by 5"? For this to be the case, N must end in 5 or 0.
(1) tells us that 0<@<4, since the factorial of any integer greater than 4 is divisible by 5. Thus, N is divisible by 5 if @=0 but is not divisible by 5 when @=1, 2, or 4. INSUFFICIENT.
(2) If @ is divisibly by 9, @=9 since no singledigit, nonzero, positive integer is divisible by any number less than itself. If @=9=units digit of N, N cannot be divisible by 5. SUFFICIENT.
Hence, my answer is B. Anyone disagree? Where am I going wrong? First of all, there is no indication about @, it could be single digit then @ = 9 then N is not a multiple of 5 it could be double digit @ = 45 then N is a multiple of 5 secondly, 0 is not divisible by 5 because 0 / 5 = 0 and Rest R =5 and we know rest must be less than quotient R<Q = 5 finally put the source of question plz



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Re: If N=1234@, is N a multiple of 5?
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18 Oct 2015, 13:49
jimwild wrote: carriedinterest wrote: If N=1234@, is N a multiple of 5? (1) @! is not divisible by 5 (2) @ is divisble by 9 The OA is C, but I can't understand why! My reasoning is as follows:
This question can be rephrased as: "is N divisible by 5"? For this to be the case, N must end in 5 or 0.
(1) tells us that 0<@<4, since the factorial of any integer greater than 4 is divisible by 5. Thus, N is divisible by 5 if @=0 but is not divisible by 5 when @=1, 2, or 4. INSUFFICIENT.
(2) If @ is divisibly by 9, @=9 since no singledigit, nonzero, positive integer is divisible by any number less than itself. If @=9=units digit of N, N cannot be divisible by 5. SUFFICIENT.
Hence, my answer is B. Anyone disagree? Where am I going wrong? First of all, there is no indication about @, it could be single digit then @ = 9 then N is not a multiple of 5 it could be double digit @ = 45 then N is a multiple of 5 secondly, 0 is not divisible by 5 because 0 / 5 = 0 and Rest R =5 and we know rest must be less than quotient R<Q = 5 finally put the source of question plz 0 is divisible by EVERY integer except 0 itself.
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Re: If N=1234@, is N a multiple of 5?
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06 Jul 2017, 09:35
If \(N = 1234@\) and @ represents the units digit, is N a multiple of 5?For \(1233@\) 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. Answer: C. OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/ifn1234an ... 02671.html
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Re: If N=1234@, is N a multiple of 5?
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