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12 May 2022, 06:45
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
42% (02:22) correct
58%
(02:25)
wrong
based on 71
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How many positive integer unordered pairs are possible such that their least common multiple and the greatest common factor are 1530 and 51, respectively?
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6
12 May 2022, 07:30
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Two numbers be
N1= 51a & N2= 51b where a&b are coprime.
Using N1*N2 = HCF *LCM
=> a*b = 30
So unordered pairs are (1,30), (2, 15), (3,10), (5,6)
Ans is 4
Posted from my mobile device
N1= 51a & N2= 51b where a&b are coprime.
Using N1*N2 = HCF *LCM
=> a*b = 30
So unordered pairs are (1,30), (2, 15), (3,10), (5,6)
Ans is 4
Posted from my mobile device
16 Jul 2025, 13:20
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Bunuel
looking at lcm and gcd we can conclude that both numbers will have 3 and 17 as common and remaining 2 , 3 , 5 can be given to both numbers in any manner. now each number from 2 ,3 , 5 has 2 options to go to so total ways we can assign them is 8. since unordered pairs are asked we divide them by 2 to avoid repeated cases. hence answer is 4(C). is my reasoning correct ?
