Question

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

Answer 1

Hint: Start by taking two different numbers and form the equation as per the statement given in the question, Apply concept of H.C.F in this equation and try to find out all the possible pairs which would satisfy both the statements.

Complete step by step answer:
Given,
Product of two numbers = 2028→(statement 1)
 HCF of those two numbers = 13→(statement 2)
Let the numbers be ‘A’ and ‘B’
Now , According to the statement 1
$A \times B = 2028 \to eqn(1)$
And according to statement 2
H.C.F = 13
Let us try to understand H.C.F first before moving further.
H.C.F stands for Highest Common Factor, which means that we are finding the largest or the highest number of common factors and factor means numbers multiplied together to form a big number.
Now , we know A and B can be represented in terms of H.C.F as
$A = 13 \times x{\text{ and }}B = 13 \times y$
Where x and y are the remaining factors of given numbers A and B respectively.
Now , Substituting these values in eqn(1) , we get
$
  A \times B = 2028 \\
   \Rightarrow (13 \times x) \times (13 \times y) = 2028 \\
   \Rightarrow 169 \times x \times y = 2028 \\
   \Rightarrow x \times y = \dfrac{{2028}}{{169}} \\
   \Rightarrow x \times y = 12 \\
    \\
$
Now , we have three possible ways in which product is 12
Case 1: 12 and 1
Which would give us
A =13×12= 156
B = 13×1 = 13
Case 2: 2 and 6
Which would give us
A =13×6 = 78
B = 13×2 = 26
Case 3: 4 and 3
Which would give us
A =13×4 = 52
B = 13×3 = 39
Product of A and B in all above cases comes out to be 2028. However, H.C.F i.e. 13 only comes out to be true for case 1 and case 3 .
Therefore, the ordered pair possible are (156,13) and (52,39)

So, the correct answer is “Option B”.

Note: Concept of L.C.M and H.C.F , Prime factorization must be very clear in order to solve such similar problems. Attention is to be given while forming the ordered pair as they should be coprime numbers.