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If N=1234@, is N a multiple of 5?

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If N=1234@, is N a multiple of 5?  [#permalink]

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New post Updated on: 06 Jul 2017, 09:35
1
9
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

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 single-digit, 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/if-n-1234-an ... 02671.html

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?  [#permalink]

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New post 04 Feb 2010, 08:30
1
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?  [#permalink]

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New post 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?  [#permalink]

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New post 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 ?

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Re: If N=1234@, is N a multiple of 5?  [#permalink]

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New post 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?  [#permalink]

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New post 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 ..!

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Re: If N=1234@, is N a multiple of 5?  [#permalink]

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New post 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 single-digit 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?  [#permalink]

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New post 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?  [#permalink]

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New post 13 Feb 2010, 12:15
2
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?  [#permalink]

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New post 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 single-digit, 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?  [#permalink]

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New post 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 single-digit, 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?  [#permalink]

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New post 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/if-n-1234-an ... 02671.html
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Re: If N=1234@, is N a multiple of 5?   [#permalink] 06 Jul 2017, 09:35
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