Question

Given two 4 digit numbers ‘abcd’ and ‘dcba’. If a+d=b+c=7, then their sum is not divisible by
$
  (a){\text{ 7}} \\
  (b){\text{ 11}} \\
  (c){\text{ 101}} \\
  (d){\text{ 111}} \\
 $

Answer 1

Hint – In this question use the concept that when two numbers are added then digits at corresponding same positions in them are added, that is unit’s place with unit’s place, tenth place with tenth place. Add digits in such a way to get the equation involving a, b, c and d. Use the information of the question to get the final value, then check for the right option that matches.

Complete step-by-step answer:
Given two four digit numbers are ‘abcd’ and ‘dcba’.
It is given that
$a + b = c + d = 7$.................... (1)
Now add the two given numbers we have,
    abcd
 + dcba
$\overline {\left( {a + d} \right)\left( {b + c} \right)\left( {c + b} \right)\left( {d + a} \right)} $
Now from equation (1) we have,
$ \Rightarrow \left( {a + d} \right)\left( {b + c} \right)\left( {c + b} \right)\left( {d + a} \right) = 7777$
So the addition of two numbers is 7777.
Now as we see that 7777 is divisible by 7, 11 and 101
$ \Rightarrow \dfrac{{7777}}{7} = 1111$, $\dfrac{{7777}}{{11}} = 707$, $\dfrac{{7777}}{{101}} = 77$
But it is not divisible by 111
As, $\dfrac{{7777}}{{111}} = 70\dfrac{7}{{111}}$
So the number formed by addition of ‘abcd’ and ‘dcba’ is not divisible by 111.
So this is the required answer.
Hence option (D) is correct.

Note – In this question when divisibility by 111 is checked the concept of mixed fraction is used. Any fraction of the form $a\dfrac{p}{q}$ depicts a mixed fraction. This mixed fraction can easily be converted into a fraction using the concept that it will be $\dfrac{{q \times a + p}}{q}$.