number properties question - logically working through a qu.

[jsl]

How do you guys work through this question logically so that you can come to a definitive answer within 2 mins? I'm struggling as I keep confusing concepts...

What is the remainder when the positive integer x is divided by 4?

1. There is no remainder when 2x is divided by 4.
2. There is no remainder when x/2 is divided by 4.

[cubicle_bound_misfit]

Let's try it

stmt1 says

2x%4 =0
==> 2x = 4k where k>0 k belongs to I

for k=1 x=2 ==> 2%4 =2 , for k=2 x=4==> 4%4 =0 not sufficient.

stmt 2 says

x/2%4=0
==> x/2 = 4k where k>0 k belongs to I

x = 8k for any value of k x%4 =0 Sufficient.

Answer B.
Please let me know OA.

Regards,

[tzink]

Just jumping in this thread, but I didn't follow that explanation..
First, what's up with the percentage signs? I assume you mean division...
Also, what do you mean "K belongs to I"?
Where does the K come from/what does it mean?

Sorry noob questions...

[cubicle_bound_misfit]

K belongs to set of all Integers

% = modulus operation i.e. 5%2 = remainder of 5 divided by 2


regards,

[target790]

(1)
2x/4 leaves no remainder where x={1,2....9}

In this scenario x can be 2,4,6 or 8

Not sufficient as there are multiple remainder options in this case.

(2)

x/8 leaves no remainder where x={1,2...9}

Only possible when x=8

So this option is sufficient.


Option B

[jsl]

Official Answer is B

[jsl]

x/8 leaves no remainder where x={1,2...9}

Thanks for this part.... can you explain why you only tested integers from 1 to 9 inclusive? Is this a general rule I can apply to all integer testing questions such as this?

[target790]

against this problem you can easily generalize.

Let me elaborate a bit more.


statement 1>x/2 where x can be any positive integer.So x should be multiples of 2(2,4,6,8.....)
In these even set you can't manage specific remainder when you divide by 4.

So this is not sufficient.


statement2>x/8 where x is any positive integers.
Now if x leaves no remainder when you divide by 8 the you can conclude easily that division by 4 will also not leave any remainder.

So this is sufficient.


Answer is B



Hope,it's now much more clear