The product of two numbers is 2028 and their H C F is 13 The number of such pairs is

let no. are 13x and 13y
13x(13y)=2028
xy=12
thus no of such pair is (1,12) and (3,4)
13 years agoHelpfull: Yes(14) No(26)
Let no.s are 13a and 13b.

so,
13a x 13b = 2028
or,
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).

Ans: there are 2 such pairs.
10 years agoHelpfull: Yes(12) No(0)
let be the numbers are 13x and 13 y
here 13x(13y)=2028
here x and y are 12.
then the number of pair is (1,12) and (3,4).
9 years agoHelpfull: Yes(3) No(4)
(6,2) is not considered because they are not co-prime number
9 years agoHelpfull: Yes(2) No(0)
given that product of numbers=2028 and hcf=13
let 13a*13b=2028
ab=2028/13*13
ab=12
if a=1, b=12 than 13,156
if a=3, b=4 than 39,52
two pairs
9 years agoHelpfull: Yes(1) No(1)
yeah why not(6,2)??
9 years agoHelpfull: Yes(0) No(0)
(6,2) is not a valid pair cause 13*6=78, 13*2=26 , for which the HCF is not 13, here they asked about the possible pairs whose HCF is 13. so (1,12),(3,4) is correct answer
8 years agoHelpfull: Yes(0) No(0)

The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is:

A. 1
B. 2
C. 3
D. 4