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If x = n + 5, y = n + 3, and x, y, and n are positive integers, what
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26 Apr 2016, 00:02
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If x = n + 5, y = n + 3, and x, y, and n are positive integers, what is the remainder when xy is divided by 6? (1) When n is divided by 6, the remainder is 3. (2) When n is divided by 3, there is no remainder.
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Re: If x = n + 5, y = n + 3, and x, y, and n are positive integers, what
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26 Apr 2016, 02:26
Bunuel wrote: If x = n + 5, y = n + 3, and x, y, and n are positive integers, what is the remainder when xy is divided by 6?
(1) When n is divided by 6, the remainder is 3. (2) When n is divided by 3, there is no remainder. xy = n^2 + 8n + 15 A) When n is divided by 6, the remainder is 3. From what's stated, n is an odd multiple of 3. n^2 = odd multiple of 3 8n = even multiple of 3= multiple of 6 15 = odd multiple of 3 so adding them together, even multiple of 3 = multiple of 6 So either A or D now coming to B) n^2 + 8n + 15 if n's a muliple of 3, when it's odd, n's an even multiple of 3. But when n's even it's an odd multiple of 3. So can't confirm if it's a multiple of 6 Answer : A
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Re: If x = n + 5, y = n + 3, and x, y, and n are positive integers, what
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26 Apr 2016, 02:53
Given: x = n + 5 y = n + 3 Remainder when xy/6 = ?
St1: n = 6k + 3 xy = (6k + 8)*(6k + 6) > 6(6k + 8)(k + 1) > Divisible by 6 Sufficient
St2: n = 3k xy = (3k + 5)*(3k + 3) = 3*(3k + 5)(k + 1) > Divisible by 6 when k is odd and not divisible by 6 when k is even Not sufficient
Answer: A



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Re: If x = n + 5, y = n + 3, and x, y, and n are positive integers, what
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26 Apr 2016, 03:27
Bunuel wrote: If x = n + 5, y = n + 3, and x, y, and n are positive integers, what is the remainder when xy is divided by 6?
(1) When n is divided by 6, the remainder is 3. (2) When n is divided by 3, there is no remainder. A : satisfies divisiblity by 6 for n=3,9,15...all B: satisfies divisibility for n=3,6,9 but not for n=0.
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Re: If x = n + 5, y = n + 3, and x, y, and n are positive integers, what
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28 Apr 2016, 21:46
Balajikarthick1990 wrote: Bunuel wrote: If x = n + 5, y = n + 3, and x, y, and n are positive integers, what is the remainder when xy is divided by 6?
(1) When n is divided by 6, the remainder is 3. (2) When n is divided by 3, there is no remainder. A : satisfies divisiblity by 6 for n=3,9,15...all B: satisfies divisibility for n=3,6,9 but not for n=0. Hi, 'N' can not be 0 as per question stem. N must be positive integer and 0 is not a positive integer. Hope this helps.



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If x = n + 5, y = n + 3, and x, y, and n are positive integers, what
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28 Apr 2016, 22:20
xy= (n+3)(n+5) for div by 6 ~~ should be div by 2 &3 statement 1: n is gives remainder =3 when div by 6 . so n+3 is div by 6. therefore xy is div by 6 sufficient
Statement 2: n is div by 3 > if n is odd multiple of 3 then n+3 is div by 6. so xy is div by 6 > id n is even multiple of 3 then (n+3) becomes odd multiple of 3 which is not div by 2 or 6 also since n is even n+5 will be odd which is again not multiple of 2. So InSUfficient
Hence A



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If x = n + 5, y = n + 3, and x, y, and n are positive integers, what
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01 Sep 2018, 20:42
Hi chetan2u PKN pushpitkc niks18 VeritasKarishmaDid Vyshak used distributive property of division ie if (a+b) is divisible by c, then a and b are EACH divisible by c, and the fact that a quotient ie k is always an integer.
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If x = n + 5, y = n + 3, and x, y, and n are positive integers, what
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01 Sep 2018, 20:57
adkikani wrote: Hi chetan2u PKN pushpitkc niks18 VeritasKarishmaDid Vyshak used distributive property of division ie if (a+b) is divisible by c, then a and b are EACH divisible by c, and the fact that a quotient ie k is always an integer. Hi adkikani, He has factored out from each of the terms in the expression 6k+6. st1: When n is divided by 6, the remainder is 3. This can be written in the form, n=6k+3 So, x=n+5=6k+3+5=6k+8 And y=n+3=6k+3+3= 6k+6=6(k+1)Hence x*y=(6k+8)*6(k+1)= 6(6k+8)*(k+1) (a*b*c=b*a*c=c*a*b) So, xy is a multiple of 6 or, xy is divisible by 6. sufficient. st2: When n is divided by 3, there is no remainder. or, n=3k+0 So, x=n+5=3k+5 And y=n+3=3k+3=3(k+1) Hence x*y=(3k+5)*3(k+1)=3*(3k+5)(k+1) a) The above expression is divisible by 3. So, xy will be divisible by 6 when either 3k+5 or k+1 is a multiple of 2, otherwise xy isn't divisible by 6. So, insufficient Hope it helps.
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Re: If x = n + 5, y = n + 3, and x, y, and n are positive integers, what
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03 Sep 2018, 04:34
adkikani wrote: Hi chetan2u PKN pushpitkc niks18 VeritasKarishmaDid Vyshak used distributive property of division ie if (a+b) is divisible by c, then a and b are EACH divisible by c, and the fact that a quotient ie k is always an integer. By the way, there is no such property. (3+4) is divisible by 7 but neither 3 is nor 4 is.
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Re: If x = n + 5, y = n + 3, and x, y, and n are positive integers, what &nbs
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