Final answer:
There is only one unique pair of numbers with the LCM of 1001 and HCF of 7, which is the pair (7, 1001).
Step-by-step explanation:
The problem asks us to find out how many pairs of numbers have the Least Common Multiple (LCM) of 1001 and the Highest Common Factor (HCF) or Greatest Common Divisor (GCD) of 7. Let's denote the two numbers by A and B. The relationship between the LCM and HCF of two numbers is given by the formula:
LCM(A, B) * HCF(A, B) = A * B
Substituting the given LCM and HCF values into the equation, we have:
1001 * 7 = A * B
7007 = A * B
Since 7007 only has prime factors 7 and 13, and we know one factor must be 7 (the HCF), the only possible pairs (A, B) that satisfy this equation are those where both A and B are divisible by 7.
Thus, A and B can take on values that are products of 7 and 13 in some combination to give us 7007 as a product. The pairs that satisfy this criterion are (7, 1001) and (91, 77).
However, as (91, 77) is essentially the same pair since they have the same prime factors, just arranged differently, we only consider distinct prime factorizations.
Therefore, there is only one unique pair of numbers with an LCM of 1001 and an HCF of 7.