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What is the remainder when the positive integer x is divided [#permalink]
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28 Oct 2010, 20:16
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What is the remainder when the positive integer x is divided by 6? (1) When x is divided by 2, the remainder is 1; and when x is divided by 3, the remainder is 0. (2) When x is divided by 12, the remainder is 3.
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Re: problem from GMAT software practice test [#permalink]
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28 Oct 2010, 21:02
mybudgie wrote: What is the best way to approach this problem? I would appreciate some help.
What is the remainder when the positive integer x is divided by 6?
Statement 1: When x is divided by 2, the remainder is 1; and when x is divided by 3, the remainder is 0.
Statement 2: When x is divided by 12, the remainder is 3. From 1... when x is devided by 3 the reminder is zero ..so it has to be multiple of 3p and the reminder is 1 when devided by 2 so it has to be odd.. 2(p) + 1 so when x is devided by 6 .. the reminder will be 1 ( when x is devisible of 6 when x is devisible of 2 and 3 both ) so 1 is suffi... Statement 2: When x is divided by 12, the remainder is 3 from when x is devided by 12 ...it has to be with 6 also...but when it carries a reminder when devided by 12... we have to think ... since the reminder we get here < 6... it has to be the same reminder when devide with the 6 also ... so the reminder will be 3 only... stmt 2 also suffi so the ans : D



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Re: problem from GMAT software practice test [#permalink]
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28 Oct 2010, 23:38
mybudgie wrote: What is the best way to approach this problem? I would appreciate some help.
What is the remainder when the positive integer x is divided by 6?
Statement 1: When x is divided by 2, the remainder is 1; and when x is divided by 3, the remainder is 0.
Statement 2: When x is divided by 12, the remainder is 3. (1) This says that x is a multiple of 3 but not a multiple of 2. All even multiples of 3 must be divisible by 6, and all odd multiples must be of the form 6n+3. (this is easy to see, consider the multiples  3,6,9,12,15,18,21,...  they are alternatively even and odd, the even ones are multiples of 6 and the others sit in the middle of two multiples, hence leave remainder 3 each time divided by 6). Sufficient
(2) x = 12k + 3 = 6(2k) + 3 Hence remainder when divided by 6 is 3 Sufficient
Answer is (d)
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Re: problem from GMAT software practice test [#permalink]
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29 Oct 2010, 02:48
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mybudgie wrote: What is the best way to approach this problem? I would appreciate some help.
What is the remainder when the positive integer x is divided by 6?
Statement 1: When x is divided by 2, the remainder is 1; and when x is divided by 3, the remainder is 0.
Statement 2: When x is divided by 12, the remainder is 3. This question can be very easily solved with number picking: (1) When x is divided by 2, the remainder is 1 > x is an odd number AND when x is divided by 3, the remainder is 0 > x is a multiple of 3 > so, x is an odd multiple of 3: 3, 9, 15, ... > any such number divided by 6 yields remainder of 3. Sufficient. (2) When x is divided by 12, the remainder is 3 > x is of a type \(x=12q+3\): 3, 15, 27, ... > any such number divided by 6 yields remainder of 3. Sufficient. Answer: D.
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Re: problem from GMAT software practice test [#permalink]
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25 Apr 2011, 17:37



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Re: problem from GMAT software practice test [#permalink]
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25 Apr 2011, 20:25
a  multiples of 3 such as 3,9,15.Each giving remainder 3. b 12 is a multiple of 6,giving remainder 3. Implies dividing the same number by 6 will give the remainder 3.
Hence OA D.



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Re: problem from GMAT software practice test [#permalink]
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26 Apr 2011, 03:47
thangduong wrote: picking number is the fastest one. using argument is not good though MORE PROFESSIONAL.
odd number which is divided by 3 is (2n+1)3, (2n+2)3 is not possible. (2n+1)3 = 6n +3 have remainder of 3 Mind you, picking numbers as a general strategy is not fool proof. Logic is. Pick numbers either as a last resort, or to get a drift of the question or when you are certain that a couple of examples will cover every scenario possible.
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Re: problem from GMAT software practice test [#permalink]
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30 Apr 2011, 17:10
best way is by picking nos in this case st 1 suff no divided by 3=== 3,6,9,12,15,18,21.... also div by 2 and remainder 1 ==== 3,9,15,21 so these no when divided by 6 will always give remainder 3 st 2 no will be 12n+3 ex 3, 15,27,39,51,... so remainder will always be 3 sufficient hence d
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Re: problem from GMAT software practice test [#permalink]
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16 Aug 2011, 09:58
Isnt the ans B. First stmt: no divided by 3=== 3,6,9,12,15,18,21.... also div by 2 and remainder 1 ==== 3,9,15,21 but when 3/6 gives reminder '0' and others give reminder 3. so this stmt is insufficient. Second Stmt: no will be 12n+3 ex 3, 15,27,39,51,... so remainder will always be 3 sufficient Or am i missing something ??? Please let me know
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Re: problem from GMAT software practice test [#permalink]
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16 Aug 2011, 10:14
g4gmat wrote: Isnt the ans B.
First stmt: no divided by 3=== 3,6,9,12,15,18,21.... also div by 2 and remainder 1 ==== 3,9,15,21 but when 3/6 gives reminder '0' and others give reminder 3. so this stmt is insufficient.
Second Stmt: no will be 12n+3 ex 3, 15,27,39,51,... so remainder will always be 3 sufficient
Or am i missing something ??? Please let me know 3/6 gives a remainder of 3 as well. The OA is D and is correct.
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