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The product of two numbers is 2028 and their HCF is 13. Find the number of such pairs.
Given :
The product of two numbers is 2028 and their HCF is 13.
To find :
We have to find the number of such pairs.
Solution :
Let the two numbers be 'a' and 'b.
The product of the two numbers is 2028.
So, $a \times b = 2028$
HCF of a and b is 13.
So, $13 \times x = a$ and $13 \times y = b$
$13x \times 13y = 2028$
$169 \times x \times y = 2028$
$x \times y = \frac{2028}{169}$
$x \times y = 12$
x and y are co-prime.
Therefore, the possible values of (x, y) are (3, 4) and (1, 12).
Then, the possible values of (a, b) are,
$13 \times 3 = 39$ and $13 \times 4 = 52$
$13 \times 1 = 13$ and $13 \times 12 = 256$
Therefore, (39, 52) and (13, 256) are such pairs of numbers.
The number of such pairs is 2.