If x = n + 5, y = n + 3, and x, y, and n are positive integers, what

Question

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.

arirux92 · Answer

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

Kurtosis · Answer

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

Balajikarthick1990 · Answer

A : satisfies divisiblity by 6 for n=3,9,15...all

B: satisfies divisibility for n=3,6,9 but not for n=0.

HarisinghKhedar · Answer

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.

Lastlap2016 · Answer

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

adkikani · Answer

Hi  chetan2u   PKN   pushpitkc   niks18   VeritasKarishma

Did  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.

PKN · Answer

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.

KarishmaB · Answer

By the way, there is no such property. (3+4) is divisible by 7 but neither 3 is nor 4 is.

KanishkM · Answer

for statement 1 to satisfy we get the cases such as 3,9,15

When substituted in x = n + 5, y = n + 3, gives a product divisible by 6

for statement 2 to satisfy, we get the cases such as 3,6,9,12,15

When substituted in x = n + 5, y = n + 3, gives a product divisible by 6 
When substituted in x = n + 5, y = n + 3, does not give a product divisible by 6

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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