Question

guys plz help me as sson as possible answer tht r not related to this questions will be reported....​

Answer 1

Answer:

13 and 156

Step-by-step explanation:

This will take long explaining and I am lazy so sorry.

Answer 2

Answer:

Please mark me as brainlest

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=2028

Also 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=

69

2028

⇒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.