Final answer:
To find two 3-digit numbers with an HCF of 45 and LCM under 3000 using the digits 0, 1, 5, 6, 7, and 8, the numbers 135 and 180 can be constructed. They both have 5 in the unit's place and share the factor 45. The LCM of these numbers is 540, fulfilling the requirement.
Step-by-step explanation:
To construct two 3-digit numbers using the digits 0, 1, 5, 6, 7, and 8 with an HCF (highest common factor) of 45 and an LCM (least common multiple) less than 3000, we would first look at the prime factorization of 45, which is 3×3×5 or 3²×5. Since 45 has to be a factor of both numbers, and both numbers are 3-digit numbers, each number must have 5 at the unit's place since zero cannot be there. Considering the LCM must be under 3000, we can use multiples of the HCF that meet this condition.
For example, we can choose 135 and 180. Both end with a 5 which aligns with the digit availability, and both include 45 as a factor. To check the LCM, we divide each number by the HCF and multiply one of the quotients with the HCF. For our example, LCM (135, 180) = HCF×[135/45]×[180/45] = 45×3×4 = 540, which is less than 3000. Thus, 135 and 180 are the two numbers we can use. Please note that there could be other combinations that also satisfy the conditions given in the problem.