The product of two numbers is 2028 and their HCF is 13 the number of such pairs is 1 2 3 4 5

Correct Answer:

Description for Correct answer:

Let the numbers be 13x and 13y where x and y are prime to each other.\( \Large \therefore LCM = 13xy \)\( \Large \therefore Producta \ of \ numbers \)\( \Large = HCF \times LCM \)\( \Large => 2028 = 13 \times 13xy \)\( \Large => xy = \frac{2028}{13 \times 13} = 12 \)

\( \Large Pairs \ satisfying \ the \ condition = \left(1, 12\right), \ and \left(3, 4\right) \)

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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

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.