The greatest number which can divide 1356, 1868 and 2764 leaving the same remainder 12 in each case is

### Concept & Formula When asked to find the greatest number that divides given numbers $x$, $y$, and $z$ leaving the same specific remainder $R$ in each case, the required number is the Highest Common Factor (HCF) of the numbers obtained by subtracting the remainder from the original numbers. Required Number = $\text{HCF of } (x - R), (y - R), \text{ and } (z - R)$ ### Step-by-Step Solution **Given:** * Numbers: $1356$, $1868$, and $2764$ * Common Remainder: $12$ **Calculation:** * Step 1: Subtract the common remainder from each of the given numbers. * $1356 - 12 = 1344$ * $1868 - 12 = 1856$ * $2764 - 12 = 2752$ * Step 2: Find the HCF of the resulting numbers ($1344, 1856, 2752$). * Step 3: Use the difference method to simplify finding the HCF. Find the differences between adjacent numbers. * $1856 - 1344 = 512$ * $2752 - 1856 = 896$ * Step 4: Find the HCF of these differences ($512$ and $896$). * $896 - 512 = 384$ * $512 - 384 = 128$ * $384 \div 128 = 3$ (Perfectly divisible) * So, the HCF of the differences is $128$. The final HCF must be a factor of $128$. * Step 5: Check the options or factors of $128$. The given numbers ($1344, 1856, 2752$) are not perfectly divisible by $128$ (e.g., $1344 \div 128 = 10.5$). The next largest factor of $128$ is $64$. * $1344 \div 64 = 21$ * $1856 \div 64 = 29$ * $2752 \div 64 = 43$ * Since $21, 29,$ and $43$ are co-prime, the HCF is definitively $64$. ### Exam Strategy & Shortcut Instead of manually calculating the HCF of large numbers, rely on the options and divisibility rules. After finding the adjusted numbers ($1344, 1856, 2752$), perform a quick scan of the choices. Since $1344 = 1300 + 44$, which isn't obviously divisible by large awkward numbers like $124$ or $156$, test the cleanest option ($64$) first. It perfectly divides all three, immediately yielding the answer. ### Common Pitfall A major mistake is calculating the HCF of the original numbers ($1356, 1868, 2764$) without subtracting the remainder first. Always remember that the remainder represents the "extra" amount preventing perfect divisibility. It must be stripped away before HCF logic applies. ### Final Answer **Therefore, the correct answer is 64.**