A number when divided by $114$, leaves remainder $21$. If the same number is divided by $19$, find the remainder.

### Concept & Logic If a number divided by $D_1$ leaves a remainder $R_1$, and we subsequently divide the same number by $D_2$ (where $D_2$ is a direct factor of $D_1$), the new remainder is found simply by dividing $R_1$ by $D_2$. ### Step-by-Step Solution **Given:** * Initial divisor ($D_1$) = $114$ * Initial remainder ($R_1$) = $21$ * New divisor ($D_2$) = $19$ **Calculation / Deduction:** * First, verify that the new divisor is a factor of the initial divisor. $114 \div 19 = 6$. Since it divides perfectly, the shortcut property holds. * Let the original number be $N$. We can express it as: $$ N = 114k + 21 $$ * Rewrite $114$ as $19 \times 6$: $$ N = 19(6k) + 19 + 2 $$ * Factor out $19$: $$ N = 19(6k + 1) + 2 $$ * This shows that when $N$ is divided by $19$, the quotient is $(6k + 1)$ and the remainder is $2$. ### Exam Strategy & Shortcut Always immediately check if the first divisor is a multiple of the second. $114$ is $19 \times 6$. Because it is a perfect multiple, you can completely ignore the first divisor. Simply take the given remainder ($21$) and divide it by the new divisor ($19$). The remainder of $21 \div 19$ is exactly $2$. ### Common Pitfall A common mistake is assuming the problem cannot be solved because the original number is unknown, prompting students to arbitrarily guess values for the quotient ($k$). While setting $k=1$ works, it wastes time compared to the direct remainder division shortcut. **Therefore, the correct answer is 2.**