ds5

[ket_gmat]

Please give explanation to attached problem

[4meonly]

I got A, but have some doubts

Question in algebraic expression:

4+7n = 3q + R, R = ?

(1)
(n+1) = 3x (multiply of 3)
4+7n = 3q + R can be expressed in
4(n+1)+3n = 3q +r
because (n+1) is divisible by 3, 4(n+1) is also divisible by 3, 3n is always divisible by 3, so R=0
SUFF

(2)
n>20
INSUFF

A

OA?

[stop@800]

4 + 7n divided by 3

3 + 6n will be divisible by 3
we are only left with (n + 1)

so r is remainder when n+1 is divided by 3


A:
Great we have
r = 0


B:
n>20
n can be anything
100000000000000000000 or 99999999
so no info


Ans IMO A

[ket_gmat]

I got A, but have some doubts

Question in algebraic expression:

4+7n = 3q + R, R = ?

(1)
(n+1) = 3x (multiply of 3)
4+7n = 3q + R can be expressed in
4(n+1)+3n = 3q +r
because (n+1) is divisible by 3, 4(n+1) is also divisible by 3, 3n is always divisible by 3, so R=0
SUFF

(2)
n>20
INSUFF

A

OA?

OA: A

[venmic]

I would do it this way

A) if (n+1) is divisble by 3 implies

n = (2,5,8...)
when 4+7n s divided by 3 then the reaminder is 0 always

Sufficent

b) n>20 imples it can be any number and can have a remainder as 0,1,2,(this is a concept)

so A