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

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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