The number of number-pairs lying between $40$ and $100$ with their H.C.F. as $15$ is

### Concept & Strategy If the H.C.F. of a pair of numbers is $15$, then both numbers must be distinct multiples of $15$. The strategy is to find all multiples of $15$ that fall strictly within the range $[40, 100]$, and then determine how many unique pairs can be formed from these multiples such that their H.C.F. remains exactly $15$. ### Step-by-Step Solution * **Given:** Range $= 40$ to $100$, $\text{H.C.F.} = 15$. * **Step 1:** List all multiples of $15$ between $40$ and $100$: * $15 \times 3 = 45$ * $15 \times 4 = 60$ * $15 \times 5 = 75$ * $15 \times 6 = 90$ So, the available numbers are: $\{45, 60, 75, 90\}$. * **Step 2:** Form all possible unique pairs from this set and verify their H.C.F.: * Pair 1: $(45, 60) \rightarrow \text{H.C.F.}(15\times3, 15\times4) = 15 \times \text{H.C.F.}(3,4) = 15$. **(Valid)** * Pair 2: $(45, 75) \rightarrow \text{H.C.F.}(15\times3, 15\times5) = 15 \times \text{H.C.F.}(3,5) = 15$. **(Valid)** * Pair 3: $(45, 90) \rightarrow \text{H.C.F.}(45, 90) = 45$. (Invalid, H.C.F. is not $15$) * Pair 4: $(60, 75) \rightarrow \text{H.C.F.}(15\times4, 15\times5) = 15 \times \text{H.C.F.}(4,5) = 15$. **(Valid)** * Pair 5: $(60, 90) \rightarrow \text{H.C.F.}(60, 90) = 30$. (Invalid, H.C.F. is not $15$) * Pair 6: $(75, 90) \rightarrow \text{H.C.F.}(15\times5, 15\times6) = 15 \times \text{H.C.F.}(5,6) = 15$. **(Valid)** * **Step 3:** Count the valid pairs: $(45, 60)$, $(45, 75)$, $(60, 75)$, and $(75, 90)$. There are $4$ valid pairs. ### Exam Strategy & Shortcut **Co-prime Factor Verification:** Instead of finding the H.C.F. of the large numbers, look at their multipliers: $\{3, 4, 5, 6\}$. A pair of numbers will have an H.C.F. of $15$ if and only if their multipliers are co-prime. Form pairs from $\{3, 4, 5, 6\}$: * $(3,4)$ - Yes * $(3,5)$ - Yes * $(3,6)$ - No (share factor 3) * $(4,5)$ - Yes * - No (share factor 2) * $(5,6)$ - Yes Total valid combinations $= 4$. This completely eliminates large-number factorization. ### Common Pitfall A frequent pitfall is assuming that *any* pair formed from the multiples of $15$ will have an H.C.F. of $15$. Students often calculate the total combinations of $4$ items taken $2$ at a time: $$\binom{4}{2} = 6$$ and incorrectly choose option (d). You must explicitly check and remove pairs like $(45, 90)$ and $(60, 90)$ where the multiplier terms share a common factor. ### Final Answer **Therefore, the correct answer is 4.**