The product of two numbers is 2028 and HCF is 13. How many possible pairs of such type?
Answer & explanation
Correct answer: option 2
Product of two factors = \(\frac{Product\;of\;two\;number}{HCF^2}\)
= \(\frac{2028}{169}\) = 12
Product of two factors possibilities:-
1, 12
3, 4
∴ 2 possibilities