Aptitude - Numbers - Discussion

Discussion :: Numbers - General Questions (Q.No.69)

69. 

n is a whole number which when divided by 4 gives 3 as remainder. What will be the remainder when 2n is divided by 4 ?

[A]. 3
[B]. 2
[C]. 1
[D]. 0

Answer: Option B

Explanation:

Let n = 4q + 3. Then 2n = 8q + 6   = 4(2q + 1 ) + 2.

Thus, when 2n is divided by 4, the remainder is 2.


Kavi said: (Mar 3, 2011)  
The simple form of these kind of sum is jus add 4+3=7 and 2n=2*7=14 which gives 2 as reminder.

Mayank Agrawal said: (Aug 25, 2015)  
What if I take n = 39(4*9+3)?

2n would be 78 and 78/4 will give me a remainder of 2.

How would you justify your answer now?

Vanitha said: (Sep 9, 2015)  
I can't understand. Can you please explain it clearly?

Goutam said: (Jul 28, 2016)  
Please explain me.

Abhishek said: (Jun 22, 2017)  
n=4q+3,
therefore adjust to n=2(2q)+2+1.
n=2(2q+1)+1............................(1).
2n=8q+6.
2n=4(2q+1)+2...........................(2).
so from 1 and 2.

Answer is 2.

Ishwariya said: (Aug 18, 2017)  
I can't able to understand. Please explain it clearly.

Gautham said: (Nov 13, 2017)  
Why it can't be like this:
n = 4q+3..... eq-01.
2n = 4q+X...eq-02.

Putting eq-01 in eq-02,
2(4q+3) = 4q+X.
8q+6 = 4q+X.
X = 4q+6.
X = 2*(2q+3).

Remainder should be 3 I don't know why they are taking 2.

Kannan said: (Jul 19, 2018)  
Here;
n reminder is 3.
2n/4=?
substitute n value 3,
2(3)/4=6/4.
6/4=2.

Naina said: (Aug 13, 2018)  
Explain it clearly?

Deepika Nair said: (Sep 6, 2018)  
I coulnt understand this concept. Could you please help me?

Shweta said: (Sep 16, 2018)  
Lt quotient be q and R be remainder.

n=4q+3 ---> eqn(1)
2n=4q+R ---> eqn(2)

therefore, putting eqn(1) in eqn (2);

2(4q+3) = 4q+R.
8q+6 = 4q + R,
let q=1,
4+6=R,
10 = R,
when 10 is divided by 4;
R=2.

Naina said: (Nov 3, 2018)  
The equation be like;

n=4*q+3 (let us assume q=1)
n=4+3
n=7.

Next equation is;
2n=4*q+x (q=1,assumed).
2(7)=4+x,
14/4 = x,
The remainder 2.

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