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A positive integer, when divided by 6 gives the remainder 2, and when
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20 Jan 2015, 05:56
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Re: A positive integer, when divided by 6 gives the remainder 2, and when
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20 Jan 2015, 08:37
Bunuel wrote: A positive integer, when divided by 6 gives the remainder 2, and when divided by 8 gives the remainder 4. What is the remainder when the integer is divided by 48?
A. 0 B. Between 1 and 6, inclusive C. Between 7 and 12, inclusive D. Between 13 and 19, inclusive E. Greater than or equal to 20
Kudos for a correct solution. n=20, 38 .. etc reminder is always going to be greater or equal to 20. Answer E.
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Re: A positive integer, when divided by 6 gives the remainder 2, and when
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21 Jan 2015, 10:04
By trying different no we can reach out to solution that is greater than or equal to 20 (E) for the question in hand
But would appreciate if someone can explain the solution through equations.
Plz help!!!!



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Re: A positive integer, when divided by 6 gives the remainder 2, and when
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21 Jan 2015, 22:27
Bunuel wrote: A positive integer, when divided by 6 gives the remainder 2, and when divided by 8 gives the remainder 4. What is the remainder when the integer is divided by 48?
A. 0 B. Between 1 and 6, inclusive C. Between 7 and 12, inclusive D. Between 13 and 19, inclusive E. Greater than or equal to 20
Kudos for a correct solution. Easiest way I know to solve this is to put in real numbers. Given number = 6n+2 = 8m+4 Start 6n+2 > 8, 14, 20, 26, 32, 38, 44, 50, 56, 62, 68.... Of these only 20 and 68 fit the description 8m+4. In both cases the remainder when divided by 48 is 20. So answer has to be E. If there is a purely algebraic way, please tell.



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Re: A positive integer, when divided by 6 gives the remainder 2, and when
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21 Jan 2015, 23:30
Bunuel wrote: A positive integer, when divided by 6 gives the remainder 2, and when divided by 8 gives the remainder 4. What is the remainder when the integer is divided by 48?
A. 0 B. Between 1 and 6, inclusive C. Between 7 and 12, inclusive D. Between 13 and 19, inclusive E. Greater than or equal to 20
Kudos for a correct solution. N = 6a + 2 Such as 2, 8, 14, 20, 26 etc N = 8b + 4 Such as 4, 12, 20, 28 etc We see that the first number that takes both these forms is 20. So smallest value of N is 20. When 20 is divided by 48, remainder will be 20. Alternatively, 6a + 2 = 8b + 4 a = (8b + 2)/6 = (4b + 1)/3 For a to be an integer, 4b+1 should be a multiple of 3. The smallest value of b for which this happens is b = 2. N = 8b+4 = 8*2 + 4 = 20 So smallest value of N is 20. When 20 is divided by 48, remainder will be 20.
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Re: A positive integer, when divided by 6 gives the remainder 2, and when
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21 Jan 2015, 23:39
82vkgmat wrote: Bunuel wrote: A positive integer, when divided by 6 gives the remainder 2, and when divided by 8 gives the remainder 4. What is the remainder when the integer is divided by 48?
A. 0 B. Between 1 and 6, inclusive C. Between 7 and 12, inclusive D. Between 13 and 19, inclusive E. Greater than or equal to 20
Kudos for a correct solution. Easiest way I know to solve this is to put in real numbers. Given number = 6n+2 = 8m+4 Start 6n+2 > 8, 14, 20, 26, 32, 38, 44, 50, 56, 62, 68.... Of these only 20 and 68 fit the description 8m+4. In both cases the remainder when divided by 48 is 20. So answer has to be E. If there is a purely algebraic way, please tell. Actually 44 is also in the list....... Did in the same way Answer = E
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A positive integer, when divided by 6 gives the remainder 2, and when
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27 Sep 2016, 11:25
Bunuel wrote: A positive integer, when divided by 6 gives the remainder 2, and when divided by 8 gives the remainder 4. What is the remainder when the integer is divided by 48?
A. 0 B. Between 1 and 6, inclusive C. Between 7 and 12, inclusive D. Between 13 and 19, inclusive E. Greater than or equal to 20
Kudos for a correct solution. Let the positive integer be N. N/6 has remainder 2 or equivalently 4 N/8 has remainder 4 or equivalently 4 Therefore, LCM of 6 and 8 = 24 will have a reminder 4 or equivalently 24  4 = 20. so, N = 24K + 20 When you divide this by 48, you will get reminder either 20 or 44. That is if N = 44 then reminder is 44, this will hold true for odd values of K. and if N = 20 then reminder is 20, this will hold true for even values of K. Hence reminder has to be greater than or equal to 20. The solution provided by Karishma is more easy to follow, but while solving the problem I recognized that that both divisions have same negative reminders so 6*8 = 48 should have also 4 remainder or 484 = 44 reminder. I was stuck there till I realized that LCM of 6 and 8 is 24 not 48.



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Re: A positive integer, when divided by 6 gives the remainder 2, and when
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17 Oct 2016, 08:08
E is correct. Here's why: From the question we can create the following equations: x=6y+2 (2,14, 20...) x=8z+4 (4, 20...) As you can see above, off these equations we can figure out various values of x based on values of z & y we feed into the equation(s). 20 is a common integer to both equations, therefore we can use it to answer the question...what is 20/48 answer is clearly 0 with a remainder of 48 and the only choice that gives us that is E



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Re: A positive integer, when divided by 6 gives the remainder 2, and when
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03 Nov 2016, 01:57
x=6q+2 x=8k+4 So let's take 20 as an example. 20 satisfies both equations. 20=0*48+20 So answer E
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Re: A positive integer, when divided by 6 gives the remainder 2, and when
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03 Nov 2016, 09:55
Bunuel wrote: A positive integer, when divided by 6 gives the remainder 2, and when divided by 8 gives the remainder 4. What is the remainder when the integer is divided by 48?
A. 0 B. Between 1 and 6, inclusive C. Between 7 and 12, inclusive D. Between 13 and 19, inclusive E. Greater than or equal to 20
Kudos for a correct solution. A = { 8 , 14 , 20 , 26 , 32...............} B = { 12 , 20 , 28 , 36............} Check the common Number it's 20...Hence, answer will be (E)
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Re: A positive integer, when divided by 6 gives the remainder 2, and when
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30 Sep 2018, 10:03
N=6p+2 so the numbers are 2,8,14,20,26,32,38 N=8q+4 so the numbers are 4,12,28,38 38 is common in both the series, Thus N becomes N = 24x+38 Now, if x=0, remainder is 38
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A positive integer, when divided by 6 gives the remainder 2, and when
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30 Sep 2018, 11:00
Bunuel wrote: A positive integer, when divided by 6 gives the remainder 2, and when divided by 8 gives the remainder 4. What is the remainder when the integer is divided by 48?
A. 0 B. Between 1 and 6, inclusive C. Between 7 and 12, inclusive D. Between 13 and 19, inclusive E. Greater than or equal to 20
Kudos for a correct solution. n=6q+2 n=8p+4 so 6q+2=8p+4 →3q4p=1 plugging in numbers, least possible value of q that gives integer value to p is 3 so least possible value of n=6*3+2=20 20/48 leaves a remainder of 20 E



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Re: A positive integer, when divided by 6 gives the remainder 2, and when
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26 Oct 2018, 04:54
Question:
the remainder remains 20. strictly speak, it can not be greater than 20, so the answer choice is worded incorrectly?
please correct me if i am wrong



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Re: A positive integer, when divided by 6 gives the remainder 2, and when
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26 Oct 2018, 05:09
Mansoor50 wrote: Question:
the remainder remains 20. strictly speak, it can not be greater than 20, so the answer choice is worded incorrectly?
please correct me if i am wrong Not necessary. 20 is the first such number. The general form of all such numbers is 24a + 20 (because 24 is the LCM of 6 and 8) So the next such number is 44 (when a = 1). If 44 is divided by 48, remainder is 44 (which is still greater than 20).
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Re: A positive integer, when divided by 6 gives the remainder 2, and when
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26 Oct 2018, 06:11
VeritasKarishma wrote: Mansoor50 wrote: Question:
the remainder remains 20. strictly speak, it can not be greater than 20, so the answer choice is worded incorrectly?
please correct me if i am wrong Not necessary. 20 is the first such number. The general form of all such numbers is 24a + 20 (because 24 is the LCM of 6 and 8) So the next such number is 44 (when a = 1). If 44 is divided by 48, remainder is 44 (which is still greater than 20). thank you!




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