The L.C.M. of two numbers is 12 times their H.C.F. The sum of H.C.F. and L.C.M. is 403. If one number is 93, find the other.

### Concept & Algebraic Logic This problem requires converting a word problem into a simple system of linear equations to isolate the H.C.F. and L.C.M., followed by applying the foundational product formula. $$ \text{First Number} \times \text{Second Number} = \text{H.C.F.} \times \text{L.C.M.} $$ ### Step-by-Step Solution * **Given Relationships:** 1. $\text{L.C.M.} = 12 \times \text{H.C.F.}$ 2. $\text{H.C.F.} + \text{L.C.M.} = 403$ 3. First number ($N_1$) = 93 * **Step 1: Find H.C.F. and L.C.M.** * Substitute the first equation into the second: * $\text{H.C.F.} + (12 \times \text{H.C.F.}) = 403$ * $13 \times \text{H.C.F.} = 403$ * $\text{H.C.F.} = \frac{403}{13} = 31$ * Now find the L.C.M.: $\text{L.C.M.} = 12 \times 31 = 372$. * **Step 2: Find the Second Number ($N_2$)** * Use the product formula: $N_1 \times N_2 = \text{H.C.F.} \times \text{L.C.M.}$ * $93 \times N_2 = 31 \times 372$ * Isolate $N_2$: $N_2 = \frac{31 \times 372}{93}$ * **Step 3: Simplify and Solve** * Notice that 93 is a direct multiple of 31 ($31 \times 3 = 93$). * Therefore, $\frac{31}{93} = \frac{1}{3}$. * $N_2 = \frac{372}{3} = 124$. ### Exam Strategy & Shortcut Avoid calculating large products. In the final step $N_2 = \frac{31 \times 372}{93}$, NEVER multiply $31 \times 372$ first. Always look for cross-cancellations vertically and diagonally. Spotting that $93 \div 31 = 3$ reduces the problem to a simple division of $372 \div 3$, saving precious time and eliminating multiplication errors. ### Common Pitfall When setting up the algebraic translation, students sometimes incorrectly write $H + 12 = 403$ instead of $H + 12H = 403$. Always pay attention to the word "times" (multiplier) versus "more than" (addition). "12 times their H.C.F." requires multiplicative scaling. ### Final Answer **Therefore, the correct answer is 124.**