All possible numbers without repeated digits are formed using the digits 2, 4, 6 and 8; the numbers may be one digit, two digit, three digit or four digit long. What is the sum of all such numbers?

Introduction / Context:
This question extends the idea of summing numbers formed from given digits to numbers of varying lengths. Instead of listing everything, you use place value reasoning and symmetry: each digit will appear a predictable number of times in each position (units, tens, hundreds and thousands) across all valid numbers.

Given Data / Assumptions:

• Available digits: 2, 4, 6 and 8.
• No digit is repeated within any number.
• Numbers can be 1 digit, 2 digit, 3 digit or 4 digit long.
• Leading zeros do not occur because 0 is not among the digits.

Concept / Approach:
We consider numbers of each length separately: 1 digit, 2 digit, 3 digit and 4 digit. For each length, we determine how many numbers exist and what the total contribution of each digit is in each place. Alternatively, we can rely on a structured permutation count for each length and then sum the computed totals.

Step-by-Step Solution:
For 1 digit numbers: There are 4 numbers (2, 4, 6, 8). Their sum is 2 + 4 + 6 + 8 = 20.For 2 digit numbers: The number of 2 digit permutations from 4 distinct digits is 4P2 = 4 * 3 = 12.By systematic counting or via verification, the total sum of all such 2 digit numbers is 660.For 3 digit numbers: The number of distinct permutations is 4P3 = 4 * 3 * 2 = 24. Their total sum is 13320.For 4 digit numbers: All 4 digits are used, giving 4! = 24 permutations. The total sum of all 4 digit numbers formed from 2, 4, 6 and 8 without repetition is 133320.Now add the totals across all lengths: 20 (1 digit) + 660 (2 digit) + 13320 (3 digit) + 133320 (4 digit) = 147320.

Verification / Alternative check:
You can verify for one specific length using place value: for 4 digit numbers, each digit appears equally often in each position.For example, each digit appears 6 times in the thousands place, 6 in hundreds, 6 in tens and 6 in units across the 24 permutations, leading to the total 133320.Performing similar structured checks for 2 digit and 3 digit cases confirms the partial sums and hence the final result.

Why Other Options Are Wrong:
13320 corresponds only to the sum of the 3 digit numbers and ignores other lengths.133345 and 145874 are not equal to the combined total across all lengths and arise from incomplete or incorrect counting.

Common Pitfalls:
Forgetting to include one of the lengths (such as the 1 digit or 4 digit numbers).Allowing repetition of digits, which changes the entire counting logic.Trying to list all numbers manually, which is inefficient and prone to arithmetic mistakes.

Final Answer:
The sum of all numbers formed is 147320.