Question

A number when divided by 143 leaves 31 as remainder. What will be the remainder when the same number is divided by 13?
a.0
b.1
c.3
d.5

Answer 1

Hint: Use the Euclid’s division lemma to solve this problem. It is expressed as a = bq + r, where ‘a’ is the number which is getting divided by number ‘b’ and hence, we get quotient as ‘q’ and remainder as ‘r’. Suppose the number as a variable and write the equations for both the conditions given in the problem and hence, try to observe the possible value of remainder.

Complete step by step answer:
As we know the Euclid’s division lemma is given as
a = bq + r …………….(i)
Where ‘a’ is the number which is getting divided by ‘b’ and hence quotient and remainder of them are ‘q’ and ‘r’ respectively.
So, as it is given in the question that a number is getting divided by 143 and it’s leaving remainder 31 and hence, we need to calculate the remainder when the same number will be divided by 13.
Now, let the number be ‘N’ and quotient be ${{Q}_{1}}$ when N is divided by 143. So, we can write an equation with help of equation (i) as
$N=143{{Q}_{1}}+31.............\left( ii \right)$
Now, let the quotient be ${{Q}_{2}}$ where the same number is getting divided by 13 and remainder left after division be R. So, we can write an equation with the help of equation (i) as
$N=13{{Q}_{2}}+R................\left( iii \right)$
Now, put the value of N from equation (i) to equation (ii) we get
$\begin{align}
  & 143{{Q}_{1}}+31=13{{Q}_{2}}+R \\
 & \Rightarrow 143{{Q}_{1}}-13{{Q}_{2}}={{R}_{2}}-31 \\
\end{align}$
Take ‘13’ as common in the left hand side of the above equation. So, we get
$13\left( 13{{Q}_{1}}-{{Q}_{2}} \right)={{R}_{2}}-31...............\left( iv \right)$
Now, we can observe equation (iv) very carefully and get that the left hand side of the expression is a multiple of 13, it means the right hand side should also be a multiple of 13. Hence, put all the values given in the option to get the value of R and observe that term R – 31 is a multiple of 13.
Option (a): 0
We get RHS as
RHS = 0 – 3 = -31
As -31 is not a multiple of 13 so, 0 is not a possible value of R
Option (b): 1
We get RHS as
RHS = 1 – 31 = -30
As -30 is not a multiple of 13 so, 1 is not a possible value of R
Option (c): 3
We get RHS as
RHS = 3 – 31 = -28
-28 is not a multiple of 13 so, 3 is not a possible value of R
Option (d): 5
We get RHS as
RHS = 5 – 31 = -26
As -6 is a multiple of 13. It means 5 is the possible value of R.
So, option (d) is the correct answer.

Note: One may write the equation
$143{{Q}_{1}}+31=13{{Q}_{2}}+R\Rightarrow 31-R=13\left( {{Q}_{2}}-13{{Q}_{1}} \right)$
Don’t get confused with the negative multiples of 13, as calculated in solution. Negative sign is there because ${{Q}_{2}}$ will be higher than $13{{Q}_{2}}$ and it will be balanced by the other side of the equations. So, don’t get confused with negative values of R – 31 in the solution.
One may think there may be other possible values of R except 5 as well, which is wrong. As we know the remainder will not be greater than 12, it can take only values from 0 to 12. So, remainder will be fixed values i.e. 5. So, if options will not be given in the problem then we need to check R for all these questions as well.