Question

guys plz help me in this question as sson as possible​

Answer 1

Answer:

2

Step-by-step explanation:

Let the two numbers be a and b

Therefore,

a×b=2028

Also, 13 is their HCF

This means that both a and b are divisible by 13

⇒

a=13×y

b=13×z where y and z are any real number

So now

ATQ

a×b=2028

⇒13y×13z=2028

⇒yz=$\frac{2028}{169}$

⇒yz=12

So possible values of y,z are (1,12), (2,6) and (3,4)

SO putting these values, we get

three pairs -> (13,156), (26,78), (39,52)

But if we see the question again,

we see that HCF must be 13

But the HCF of the second pair is 26 not 13

So there are only two pairs

Answer 2

Answer:

2 pairs

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 there HCF,thus both numbers must be divisible by 13.

So, Let x = 13a and y = 13b, then

13a ×13b = 2028

169ab = 2028

ab = 2028 ÷ 169

ab = 12

Therefore, required possible pair of values of x and y which are prime to each other are

( 1,12) & (3,4).

Thus the required numbers are ( 12 , 156 ) & ( 39 , 52 ).

Hence, the number of possible pair is 2.

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Hope this answer help you

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