Question

The difference of three-digit number and the number
obtained by putting the digits in reverse order is always
divisible by
a) 4and 3
b) 4and 9
c)11and 9
d( 9 only​

Answer 1

Question:

The difference of 3-digits number and the number obtained by putting the digits in reverse order is always divisible by :

a) 4 and 3

b) 4 and 9

c) 11 and 9

d) 9 only

Answer:

c) 11 and 9

Note:

• Rule of divisibility by 3 : If the sum of all the digits of a number is divisible by 3, then the number itself will also be divisible by 3.

• Rule of divisibility by 9 : If the sum of all the digits of a number is divisible by 9, then the number itself will also be divisible by 9 .

• Rule of divisibility by 11 : If the difference between the sum of the digits at the even places and the sum of the digits at odd places is divisible by 11, then the number itself will also be divisible by 11 .

Solution:

Let N be a 3-digits number as ;

$\begin{tabular}{|c|c|c|}\cline{1-3} Hundreds digit & Tens digit & Unit digit \\\cline{1-3}x&y&z \\\cline{1-3}\end{tabular}$

Thus ,

The number N will be given as ;

N = 100x + 10y + z

Now,

Let N' be the number obtained after reversing the digits.

$\begin{tabular}{|c|c|c|}\cline{1-3} Hundreds digit & Tens digit & Unit digit \\\cline{1-3}z&y&x \\\cline{1-3}\end{tabular}$

Also,

The number N' will be given as ;

N' = 100z + 10y + x

Now,

Let D be the between the original number and the number obtained after reversing its digits.

Thus,

=> D = | N - N' |

=> D = | ( 100x+10y+z ) - ( 100z+10y+x ) |

=> D = | 100x + 10y + z - 100z - 10y - x |

=> D = | 99x - 99z |

=> D = 99•| x - z |

From the above equation, it is clear that ,

99•| x - z | is divisible by 3 , 9 and 11 , thus the difference D is divisible by 3 , 9 and 11 .

Hence,

The right answer from the given options is : c) 11 and 9 .

Example :

Let a digits number be 123.

Then ,

The number obtained after reversing its digits will be 321.

Thus,

The difference between 123 and 321

= | 123 - 321 |

= | - 198 |

= 198

Test of divisibility of 198 by 3 and 9 :-

Sum of digits = 1 + 9 + 8 = 18

Clearly,

The sum of digits (ie, 18) is divisible by 3 and 9 .

Thus,

198 ( which is the difference between the number and the number obtained by reversing its digits ) is divisible by 3 and 9 .

Test of divisibility of 198 by 11 :-

Sum of digits at odd places = 1 + 8 = 9

Sum of digits at even places = 9

Now,

The difference between the sum of digits at odd places and the sum of digits at even places = 9 - 9 = 0

Clearly,

The difference between the sum of digits at odd places and the sum of digits at even places (ie , 0) is divisible by 11.

Thus,

198 ( which is the difference between the number and the number obtained by reversing its digits ) is divisible by 11 .

Hence,

The difference between a 3-digits number and the number obtained by reversing its digits is always divisible by 3 , 9 and 11 .