### Discussion :: Problems on Numbers - General Questions (Q.No.8)

8. | A number consists of two digits. If the digits interchange places and the new number is added to the original number, then the resulting number will be divisible by: |
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Answer: Option D Explanation: Let the ten's digit be Then, number = 10 Number obtained by interchanging the digits = 10 (10 Video Explanation: https://youtu.be/lytJE8GqRvM |

Farukh said: (Jun 8, 2011) | |

I didn't get it. Can anyone explain ? Please. |

Nikunjjain said: (Aug 21, 2011) | |

Why did they take ten digits number in the solution of question because in question they have mentioned 2 digits no ? |

Krishna said: (Sep 9, 2011) | |

Two digits No. means any of the number from 10 to 99. let us take 98 as a 2 digit number. In this number 9 tens and one 8 is there. 9 tens means 9x10. that is 90 and one 8 is being added. So that we can get 98. That is why they have mentioned as ten's digit. for your understanding we will take a 98743265. In this number 5 is in the place of one, 6 is 10th, 2 is 100th, 3 is 1000th, 4 is 10,000th,7 is 1,00,000th 8 is 10,00,000th and 9 is 1,00,00,000th. |

Suja said: (Sep 30, 2011) | |

Wow karishma yaar you are so talented sahi samzaya answer. Now I got it. |

Santhosh said: (Oct 31, 2011) | |

It's very simple.. for example take a two digit number : i.e 45 the reverse of the number is 54 So 45+54= 99(which is a divisible of 11) |

Virendra said: (Nov 22, 2011) | |

Like santosh it is really simple just start by taking 12....21., at first attempt option 5 and 9 cancelled out at second attempt 13....31 we get ans is 11. |

Sarika said: (Apr 20, 2012) | |

Take any two numbers. For ex take 63+36=99. So 99 is divisible by 11. If we take any two numbers all that numbers are only divisible by 11 only. So answer is 11. |

Hemanth said: (Jul 15, 2013) | |

It is very easy, Check with answer, in answer two-digit number is 11. Then interchange it will '11'. Then add old and new is = 11+11 = 22. Sum:22, check with choice 22 is divisible by 11. |

Ashish Singh said: (Sep 9, 2014) | |

Very easy yar. Take 10,11,......99 any number. Just interchange position then add it, Say 23+32 = 55. 24+42 = 66. 10+01 =11. 45+54 = 99. |

Sajid said: (Sep 10, 2014) | |

10 to 99. 10, 11, 12, 13, 14, 15, 16, 17....... 10+01=11; 12+21=33; 13+31=44; 14+41=55. Eg: 28+82=110; Any two numbers from 10 to 99 gets interchange of 10 unit position to one unit position and one unit position to 10 unit position and do sum of the numbers that particular number will be definitely divisible 11. |

Vinay said: (Oct 4, 2015) | |

When Should we take a 2 digit as 10x+y. And when should we take as x+y? Anybody help? |

Kunal said: (Nov 21, 2015) | |

If we take the no as 99. After reversing the no stays as 99. Then new no added to original no, we get 99+99 = 198. 198 is divisible by 9 as well as 11. So can anyone help out about this doubt. |

Raja said: (Jan 25, 2016) | |

Simple answer. 10 = x. y = "1" for example. 10x+y = 11. 11 is a answer. So option is D. |

Santosh H said: (Mar 8, 2016) | |

11 = 10x+y = 10(1)+1. Another ex: 25=10x+y = 10(2)+5. Also: 255=100x+10y+z = 100(2)+10(5)+5. |

Yogita said: (Oct 17, 2016) | |

@Santhosh. I'm pretty confused that when we have to take x, y, and when it is only x for the formation of two digit number? |

Pooja P Kumar said: (Apr 18, 2017) | |

@Yogita. Whenever the product of two numbers is asked. For example: The product of two number is such that the result is 5.. so we will take same variables i.e x because x*5/x is 5. And mostly in all other cases, we will take two variables x and y. |

Annah said: (Feb 20, 2018) | |

(10x+y) + (10y + x)= For; (10x+y) Let x be 2and y be 1, substitute in 10x +y, (10*2) + 1=21. For; (10y+ x) interchange. Let x be 2 and y be 1, substitute in (10y+x), (10*1)+ 2=12. Then, 12+ 21=33 which is divisible by 11. |

Ramya Sana said: (Aug 1, 2018) | |

For example if we take 27 and when we reverse it, we will get 72 and addition of these two numbers is 99 which will also divisible by 3 and 9 and 11. Then what is the answer. Anybody can help me out? |

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