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Question Stats:
34% (01:34) correct 66% (01:21) wrong based on 183 sessions
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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
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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: D0112
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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 1234 to 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
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Re: D0112
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22 Oct 2014, 23:50
lastshot wrote: 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 1234 to 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: D0112
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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 1234 to be divisible by 9, @ must be 8. Answer choice B is enough to answer this question.



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Re: D0112
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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 1234 to 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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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.
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Re: D0112
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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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Re: D0112
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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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Re: D0112
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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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Re D0112
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05 May 2016, 08:43
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.



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Re: D0112
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05 May 2016, 08:50
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 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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Re: D0112
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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
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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
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09 Aug 2016, 01:43
mounirbr wrote: Hello, regardin the first condition, why @ can't be 6 or 7? both ! are not divisible by 5.
thanks 6! = 1*2*3*4* 5*6 = 720, which IS divisible by 5; 7! = 1*2*3*4* 5*6*7 = 5040, which IS divisible by 5.
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Re: D0112
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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
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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
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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
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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
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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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