A number when divided by $136$ leaves remainder $36$. If the same number is divided by $17$, the remainder will be

### Concept & Rule This is a classic successive divisor problem. If a number $N$ divided by divisor $D_1$ leaves remainder $R_1$, and you are asked for the remainder when $N$ is divided by $D_2$, simply check if $D_1$ is a multiple of $D_2$. If it is, the new remainder is just $R_1$ divided by $D_2$. $$ N = (D_1 \times Q) + R_1 $$ ### Step-by-Step Solution * **Given:** * Original Divisor ($D_1$) = $136$ * Original Remainder ($R_1$) = $36$ * New Divisor ($D_2$) = $17$ * **Calculation:** * First, verify that the first divisor ($136$) is completely divisible by the second divisor ($17$). * $136 / 17 = 8$ (Yes, it is perfectly divisible). * Because $136$ is a multiple of $17$, the "$136 \times Q$" portion of the number will always divide perfectly by $17$. * Therefore, we only need to divide the original remainder by the new divisor. * $36 / 17 = 2$ with a remainder of $2$. ### Exam Strategy & Shortcut Do not write out the equations. The moment you see this problem format, quickly mentally check $136 / 17$. Once confirmed it divides evenly, immediately calculate $36 / 17$. The remainder is $2$. This should take no more than $5$ seconds in an exam. ### Common Pitfall Students sometimes try to invent a value for the original number by assuming a quotient of $1$ (i.e., $136 \times 1 + 36 = 172$), and then dividing $172$ by $17$. While this works and yields the correct answer, it costs valuable time calculating larger numbers unnecessarily. ### Final Answer Therefore, the correct answer is **2**.