The H.C.F. of two numbers, each having three digits is $17$ and their L.C.M. is $714$. Find the sum of the numbers.

### Concept & Formula If the H.C.F. of two numbers is $H$, the numbers can be represented as $Hx$ and $Hy$, where $x$ and $y$ are co-prime integers. Furthermore, the product of two numbers is equal to the product of their H.C.F. and L.C.M. $$ \text{Product of Numbers} = \text{H.C.F.} \times \text{L.C.M.} $$ $$ (Hx) \times (Hy) = H \times L $$ ### Step-by-Step Solution * **Given:** H.C.F. $= 17$, L.C.M. $= 714$. Both unknown numbers are strictly 3-digit numbers. * Let the two numbers be $17x$ and $17y$, where $x$ and $y$ are co-prime. * Using the property: $17x \times 17y = 17 \times 714$ * $289xy = 12138$ * $xy = \frac{12138}{289} = \frac{714}{17} = 42$ * Find pairs of co-prime numbers $(x, y)$ that multiply to $42$: * $(1, 42)$ * $(2, 21)$ * $(3, 14)$ * $(6, 7)$ * Since the original numbers ($17x$ and $17y$) must be 3-digit numbers, their values must be greater than $99$. * If $x = 1$, the number is $17 \times 1 = 17$ (2-digit, rejected). * If $x = 2$, the number is $17 \times 2 = 34$ (2-digit, rejected). * If $x = 3$, the number is $17 \times 3 = 51$ (2-digit, rejected). * If $x = 6$, the number is $17 \times 6 = 102$ (3-digit, valid). * If $y = 7$, the number is $17 \times 7 = 119$ (3-digit, valid). * The two numbers are $102$ and $119$. * Their sum is $102 + 119 = 221$. ### Exam Strategy & Shortcut Instead of testing all pairs mentally, just look at the condition "three digits". $17 \times 5 = 85$, so you need multipliers strictly $6$ or higher. The only factors of $42$ where both are $\geq 6$ are $6$ and $7$. Directly compute the sum as $17 \times (6 + 7) = 17 \times 13 = 221$. ### Common Pitfall A common error is stopping at $xy = 42$ and arbitrarily picking $(1, 42)$ or $(2, 21)$ as the pair, leading to completely incorrect 2-digit numbers. Always cross-verify the specific constraints given in the problem (e.g., "three digits"). ### Final Answer **Therefore, the correct answer is 221.**