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An arithmetic sequence is a series of numbers with a constant difference between consecutive terms. In the given exercise, we solve problems involving multiples of specific numbers using arithmetic sequences. For example, the sequence of numbers divisible by 7 less than 1000 can be described by the sequence: 7, 14, 21, ..., where each number increases by 7. To find the number of these terms, we use the formula for the nth term of an arithmetic sequence: \(a_n = a + (n-1)d\), where \(a\) is the first term and \(d\) is the common difference. Setting the nth term to be less than 1000, we solve for the integer value of \(n\), leading to \(n = \left\lfloor \frac{1000-7}{7} \right\rfloor = 142\). This tells us there are 142 multiples of 7 under 1000.

The least common multiple (LCM) of two numbers is the smallest number that is a multiple of both. In the exercise, we compute the LCM to determine how many numbers are divisible by both 7 and 11 under 1000. The approach for the LCM of 7 and 11 involves finding a number that both 7 and 11 evenly divide. Since 7 and 11 are coprime (have no common factors other than 1), the LCM is simply their product: \(7 \times 11 = 77\). To find how many numbers are divisible by this LCM (77) under 1000, we use the formula mentioned above: \( \left\lfloor \frac{999}{77} \right\rfloor = 12\). Thus, there are 12 numbers divisible by both 7 and 11 less than 1000.

The principle of inclusion-exclusion is a counting technique used to calculate the size of the union of multiple sets. It helps avoid overcounting elements that are in the intersection of the sets. For the problem, to find how many numbers under 1000 are divisible by either 7 or 11, we start by calculating those divisible by each individually: 142 multiples of 7 and 90 multiples of 11. Using the inclusion-exclusion principle, we subtract those counted twice (multiples of both 7 and 11): \(142 + 90 - 12 = 220\). Therefore, there are 220 numbers under 1000 that are divisible by either 7 or 11.

Numbers with distinct digits do not repeat any digit within them. For example, the number 123 has distinct digits, but 121 does not. In counting such numbers under 1000, one-digit and two-digit numbers inherently have distinct digits. For three-digit numbers, the first digit can be any from 1 to 9, the second any digit except the first (9 choices), and the third except the other two (8 choices), giving: \(9 \times 9 \times 8 = 648\). Including all possibilities, the total is: 9 (one-digit) + 90 (two-digit) + 648 (three-digit) = 747 numbers. To count those with distinct digits and even, the units place must be 0, 2, 4, 6, or 8, and adjust choices for other digits accordingly. The calculations are more intricate, but the principle remains ensuring non-repetition and verifying end-even criteria.