The sum of two numbers is $528$ and their H.C.F. is $33$. The number of pairs of numbers satisfying the above conditions is

### Concept & Formula Let the two numbers be $33a$ and $33b$, where $a$ and $b$ are strictly **co-prime** integers ($\text{H.C.F.}(a, b) = 1$). Using the sum condition: $$33a + 33b = \text{Sum}$$ The number of valid co-prime pairs $(a, b)$ determines the total number of pairs of numbers that satisfy the given conditions. ### Step-by-Step Solution * **Given:** Sum $= 528$, $\text{H.C.F.} = 33$. * **Step 1:** Represent the numbers as $33a$ and $33b$. * **Step 2:** Formulate the equation: $$33a + 33b = 528$$ $$33(a + b) = 528$$ * **Step 3:** Isolate $(a + b)$: $$a + b = \frac{528}{33} = 16$$ * **Step 4:** List all positive integer pairs $(a, b)$ such that $a + b = 16$, and filter for co-prime pairs: * $(1, 15) \rightarrow$ Co-prime ($\text{H.C.F.} = 1$). **(Valid)** * $(2, 14) \rightarrow$ Both even ($\text{H.C.F.} = 2$). (Invalid) * $(3, 13) \rightarrow$ Co-prime ($\text{H.C.F.} = 1$). **(Valid)** * $(4, 12) \rightarrow$ Share factor $4$. (Invalid) * $(5, 11) \rightarrow$ Co-prime ($\text{H.C.F.} = 1$). **(Valid)** * $(6, 10) \rightarrow$ Both even ($\text{H.C.F.} = 2$). (Invalid) * $(7, 9) \rightarrow$ Co-prime ($\text{H.C.F.} = 1$). **(Valid)** * $(8, 8) \rightarrow$ Share factor $8$. (Invalid) * **Step 5:** Count the valid pairs: $(1, 15)$, $(3, 13)$, $(5, 11)$, and $(7, 9)$. There are exactly $4$ pairs. ### Exam Strategy & Shortcut **Coprimality Count:** Once you find $a + b = 16$, you can find the number of valid pairs by calculating how many numbers less than $16$ are co-prime to $16$ (Euler's totient function $\phi(16)$) and dividing by $2$ (since order doesn't matter for pairs). Since $16 = 2^4$, the numbers not co-prime to it are all even numbers. The odd numbers less than $16$ are $1, 3, 5, 7, 9, 11, 13, 15$ (8 numbers). Pairs $= 8 / 2 = 4$. This avoids listing and checking every single pair manually. ### Common Pitfall A very common mistake is counting all possible additive combinations that sum to $16$ (which would give $8$ pairs, leading to option c). Always remember to filter out any pairs that share a common factor, as their inclusion would change the H.C.F. from $33$ to a higher multiple. ### Final Answer **Therefore, the correct answer is 4.**