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D01-12

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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
[Reveal] Spoiler: OA

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Official Solution:

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

\(1234@\) 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
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Re: D01-12 [#permalink]

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New post 22 Oct 2014, 22:38
Bunuel wrote:
Official Solution:


\(1234@\) 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



I am not getting one point Bunnel !!!

As per statement B : 1234@ is divisible by 9, that means sum of digits must also be divisible by 9 (As per divisibility rules)

Sum of digits is 1+2+3+4 = 10, thus in order for 1234to be divisible by 9, @ can have value of 8 only.

Which makes sum equal to 18, & complete number to 12348,

Thus solution must be B.

Shade some light , if i am missing any point !!!

Regards
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Bunuel wrote:
Official Solution:


\(1234@\) 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



I am not getting one point Bunnel !!!

As per statement B : 1234@ is divisible by 9, that means sum of digits must also be divisible by 9 (As per divisibility rules)

Sum of digits is 1+2+3+4 = 10, thus in order for 1234to be divisible by 9, @ can have value of 8 only.

Which makes sum equal to 18, & complete number to 12348,

Thus solution must be B.

Shade some light , if i am missing any point !!!

Regards
LS


(2) says that the units digit, \(@\), is divisible by 9, not the number itself.
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Re: D01-12 [#permalink]

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New post 27 Nov 2014, 10:21
As per statement B : 1234@ is divisible by 9, which means sum of digits must be divisible by 9

Sum of 1+2+3+4 = 10, therefore 1234to be divisible by 9, @ must be 8. Answer choice B is enough to answer this question.

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New post 28 Nov 2014, 07:20
nice question, i assumed the whole number is divisible by 9... my bad!!!! :(
thanks @Bunnuel

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New post 28 Nov 2014, 08:27
gompgo wrote:
As per statement B : 1234@ is divisible by 9, which means sum of digits must be divisible by 9

Sum of 1+2+3+4 = 10, therefore 1234to be divisible by 9, @ must be 8. Answer choice B is enough to answer this question.


The correct answer is C as given in the solution above as well in the original post under the spoiler.

(2) says that the units digit, \(@\), is divisible by 9, not the number itself.
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New post 05 Dec 2014, 05:02
Nice question with subtle logic..+1 for Bunuel. :)
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Is there any Property that if a number Q is not Divisible by 5 then Q!(factorial) is also not divisible by 5??

Ex: If Q = 14 not Divisible by 5, then 14! ( 14x13x12....1) Can't we consider that 10 ,5 are in the factorial list so 14! can be divisible.
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New post 06 Dec 2014, 04:40
I totally forgot to consider 0! That's frustrating!

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New post 06 Dec 2014, 08:22
kanusha wrote:
Bunnel
Is there any Property that if a number Q is not Divisible by 5 then Q!(factorial) is also not divisible by 5??

Ex: If Q = 14 not Divisible by 5, then 14! ( 14x13x12....1) Can't we consider that 10 ,5 are in the factorial list so 14! can be divisible.


How this could be correct? If q=14, then q! = 1*2*3*4*5*6*7*8*9*10*11*12*13*14 IS a multiple of 5.

Generally, for any integer n more than or equal to 5, n! is divisible by 5 because it will have 5 as a multiple.
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D01-12 [#permalink]

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New post 01 Mar 2015, 06:31
Bunuel wrote:
Official Solution:

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

\(1234@\) 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






Hi,

I got the answer A

As @ can be either 0 or 5 (multiple of 5)

(A) states that @! is not a multiple of 5
Hence, 5! is a multiple of 5
and 0! that is 1, is not a multiple of 5

So, shouldn't be 0?

Please help

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New post 21 Oct 2015, 22:40
I don't agree with the explanation. for a number to be divisible by 9, it is mandatory that sum of all its digits adds up to 9. Here we have the number as 1234X.
So the sum of existing digits is 1+2+3+4=1, hence X can be 8 for the number to be divisible by 9. Hence the explanation for statement b is not true.

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New post 21 Oct 2015, 22:51
apurva297 wrote:
I don't agree with the explanation. for a number to be divisible by 9, it is mandatory that sum of all its digits adds up to 9. Here we have the number as 1234X.
So the sum of existing digits is 1+2+3+4=1, hence X can be 8 for the number to be divisible by 9. Hence the explanation for statement b is not true.


Please read carefully. (2) says that @ is divisible by 9 not N.
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New post 18 Dec 2015, 06:27
I think this is a high-quality question and I agree with explanation.
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New post 25 Dec 2015, 03:17
I just have one question.

explanation says that 0 is divisible by every integer, can anyone elaborate on that?

Because I think 0 is not divisible by any integer.

Kindly help.

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New post 25 Dec 2015, 03:22
neha1993 wrote:
I just have one question.

explanation says that 0 is divisible by every integer, can anyone elaborate on that?

Because I think 0 is not divisible by any integer.

Kindly help.

We know that 0/2 = 0. Right?

0/2 means 2 divides 0 - 0 times.
0/5 means 5 divides 0 - 0 times.

In short, all numbers divide 0 - 0 times. This is how all numbers divide 0. :)

+Kudos, if this helped.
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New post 06 Apr 2016, 07:45
I think this is a high-quality question and I agree with explanation.

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New post 05 May 2016, 08:43
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.

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New post 05 May 2016, 08:50
amilbusthon wrote:
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 question is fine. You just did not read it carefully: (2) says that the units digit, @, is divisible by 9, not the number itself.
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