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.

Difficulty: Medium

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

More Questions from Problems on H.C.F and L.C.M

Discussion & Comments

No comments yet. Be the first to comment!
Join Discussion