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# A number when divided by $114$ leaves the remainder$21$. Then find the remainder if the same number is divided by $19$ .A)$1$ B)$2$ C) $7$ D) $21$  Verified
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Hint: We know the formula ${\text{Dividend = Divisor}} \times {\text{Quotient + Remainder}}$.We can put the values given in first statement in this formula. Then we can write $114 = 19 \times 6$ and $21 = 19 + 2$.When we adjust the equation, we can find the remainder.

Let the number which is dividend be x. Also given a second divisor=$19$ and if x is divided by this number we have to find the remainder. Quotient is not given so let us denote is by Q. Now, we know that,
${\text{Dividend = Divisor}} \times {\text{Quotient + Remainder}}$
When we put the values of first divisor and remainder we get,
$\Rightarrow {\text{x = 114}} \times {\text{Q + 21}}$
Now we have to find the remainder for the second divisor and we have two unknown quantities here. So we will change the values of formula so that we can find the remainder. Now we can write $114 = 19 \times 6$ and $21 = 19 + 2$ because we have to change the divisor from $114$ to $19$ so we can find the remainder. Then the equation will change into-
$\Rightarrow$ ${\text{x = }}\left( {19 \times 6} \right) \times {\text{Q + }}\left( {{\text{19 + 2}}} \right)$
$\Rightarrow {\text{x = 19}} \times \left( {{\text{Q}} \times {\text{6}}} \right){\text{ + 19 + 2}}$
On taking $19$ common, we get
$\Rightarrow {\text{x = 19}}\left( {{\text{Q}} \times {\text{6 + 1}}} \right){\text{ + 2}}$
This means that the Divisor =$19$ and Quotient=$\left( {{\text{Q}} \times {\text{6 + 1}}} \right)$ and the remainder is $2$
$\Rightarrow \left( {\dfrac{{{\text{given remainder}}}}{{{\text{second divisor}}}}} \right)$
$\Rightarrow$ $\dfrac{{21}}{{19}}$ $= 1$ and remainder left is $2$ . So $2$ is the required remainder.