The product of two numbers is 2028 and their HCF is 13 how many such pairs are there

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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

Verified

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

Let the two numbers are a and b. Now their H.C.F is given as 13. So this means 13 is the greatest common divisor of both of them.So we can write \[a=13{{k}_{1}}\] and \[b=13{{k}_{2}}\] where \[{{k}_{1}},{{k}_{2}}\] are positive integers and H.C.F of \[{{k}_{1}},{{k}_{2}}\] is 1.Now it is given that, ab = 2028So putting the values of a and b we get,\[(13{{k}_{1}})\cdot (13{{k}_{2}})=2028\]\[\Rightarrow 169{{k}_{1}}{{k}_{2}}=2028\]Alternatively, we can write \[{{k}_{1}}{{k}_{2}}=12\]Now 12 can be written as a product of two positive integers where they are 12’s divisors.We can write 12 as a product of 1 and 12 or 2 and 6 or 3 and 4 where order does not matter.Now among the above three pairs, H.C.F of 2 and 6 is 2 which is not 1. Therefore, we can discard it.The remaining two pairs are (1, 12) and (3, 4) respectively.This can be our required values of \[{{k}_{1}},{{k}_{2}}\]If we take \[{{k}_{1}}=1\] and \[{{k}_{2}}=12\] then we get \[a=13\] and \[b=156\]Again if we take \[{{k}_{1}}=3\] and \[{{k}_{2}}=4\] then we get \[a=39\] and \[b=52\]Hence, we get two pairs of numbers, which satisfies the given conditions. These are (13,156) and (39,52) respectively.

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.