guys plz help as sson as possible answer tht is not related to this quetion will eb deleted and rwported
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the number of such pair is 2
Step-by-step explanation:
Let the two numbers be x and y respectively.It is given that the product of the two numbers is 2028, therefore, xy=2028Also 13 is their HCF, thus both numbers must be divisible by 13.So, let x=13a and y=13b, then 13a×13b=2028⇒169ab=2028⇒ab= 692028 ⇒ab=12 Therefore, required possible pair of values of x and y which are prime to each other are (1,12) and (3,4).Thus, the required numbers are (12,156) and (39,52).Hence, the number of possible pairs is 2.
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