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249
Q:
Answer: B) 2
Explanation:
Let the numbers 13a and 13b.
Then, 13a x 13b = 2028
=>ab = 12.
Now, the co-primes with product 12 are (1, 12) and (3, 4).
[Note: Two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]
So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4).
Clearly, there are 2 such pairs.
Subject: HCF and LCM - Quantitative Aptitude - Arithmetic Ability
Answer
Hint: Consider the two number as a and b. So, we have ab= 2028 and the greatest divisor of them is 13. Write a and b in multiples of 13 and solve for the answer.
Complete step-by-step solution
Hence, the correct option to this question is option (b) 2.
Note: As H.C.F of a and b is 13 and already taken out then the H.C.F of \[{{k}_{1}},{{k}_{2}}\] must be 1 in order to remain the H.C.F of a and b, 13. As the product is commutative so the pair (a,b) and (b, a) are equivalent and hence have not counted twice.
The product of two numbers is 2028 and their HCF is 13. The number of such pairs is : [A]1 [B]2 [C]3 [D]4
2 Here, HCF = 13 Let the numbers be 13x and 13y, where x and y are prime to each other. Now, 13x $latex \times$ 13y = 2028 $latex => xy = \frac{2028}{13\times 13} = 12&s=1$ The opposite pairs are : (1, 12), (3, 4), (2, 6) But the 2 and 6 are not co-prime. So, required no. of pairs = 2
Hence option [B] is correct answer.