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The sum of two non co–prime numbers added to their HCF gives us 91. How many such pairs are possible?
- a)2
- b)4
- c)3
- d)6
- e)8
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
The sum of two non co–prime numbers added to their HCF gives us ...
Let HCF of the numbers be h. The numbers can be taken as ha + hb, where a, b are coprime.
h + ha + hb = 91
h(1 + a + b) = 91
h ≠ 1
h = 7
=> 1 + a + b = 13 a + b = 12
h + ha + hb = 91
h(1 + a + b) = 91
h ≠ 1
h = 7
=> 1 + a + b = 13 a + b = 12
h = 13
=> 1 + a + b = 7
=> a + b = 6
=> 1 + a + b = 7
=> a + b = 6
Case 1: h = 7, a + b = 12
(1, 11), (5, 7) => Only 2 pairs are possible as a, b have to be coprime.
(1, 11), (5, 7) => Only 2 pairs are possible as a, b have to be coprime.
Case 2: h = 13, a + b = 6
(1, 5) only one pair is possible as a, b have to be coprime.
(1, 5) only one pair is possible as a, b have to be coprime.
Overall, 3 pairs of numbers are possible – (7, 77) (35, 49) and (13, 65)
The question is "How many such pairs are possible?"
Hence the answer is "3 Pairs"
Most Upvoted Answer
The sum of two non co–prime numbers added to their HCF gives us ...
Mplex numbers is also a non complex number.
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