Two numbers have product 7168 and highest common factor (HCF) 16. Considering all possible such pairs, find the sum of all the distinct numbers involved.
Aptitude
Problems on H.C.F and L.C.M
Difficulty: Medium
Choose an option
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A640
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B860
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C460
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DData inadequate
Answer
Correct Answer: 640
Explanation
Introduction / Context:When product and HCF are known, we can parameterize the two numbers as multiples of the HCF with co-prime multipliers. This approach lets us enumerate all valid unordered pairs and then compute the requested sum over distinct values appearing in those pairs.
Given Data / Assumptions:
- a * b = 7168.
- HCF(a, b) = 16.
- a, b are positive integers.
Concept / Approach:Let a = 16x and b = 16y with HCF(x, y) = 1. Then 256xy = 7168 ⇒ xy = 7168 / 256. Factor the right-hand side, and list the coprime factor pairs (x, y). Each gives a valid unordered pair (a, b). Finally, add all distinct numbers obtained from these pairs.
Step-by-Step Solution:
Compute xy: 7168 / 256 = 28.Coprime factor pairs of 28: (1, 28) and (4, 7). (2, 14) is not coprime.Corresponding (a, b) pairs: (16*1, 16*28) = (16, 448) and (16*4, 16*7) = (64, 112). Orders reversed are the same unordered pairs.Distinct numbers appearing: 16, 448, 64, 112. Sum = 16 + 448 + 64 + 112 = 640.Verification / Alternative check:
Check HCFs: HCF(16, 448) = 16; HCF(64, 112) = 16. Products: 16*448 = 7168 and 64*112 = 7168, both correct.Why Other Options Are Wrong:
- 860, 460 do not match the computed sum of distinct elements across all valid pairs.
- “Data inadequate” is incorrect because the parameterization produces a finite, explicit set of pairs.
Common Pitfalls:
- Including non-coprime pairs like (2, 14) which would make the HCF greater than 16.
Final Answer:
640