Sum of all 4-digit numbers using digits 2, 4, 6, 8 with repetitions allowed:

To find the sum of all 4-digit numbers that can be formed using the digits 2, 4, 6, and 8 with repetitions allowed, we need to consider all possible combinations of these digits in the thousands, hundreds, tens, and units places.

1. Permutations:
The first step is to find all the possible permutations of the given digits. Since repetitions are allowed, we can have four options for each place value. The total number of permutations can be calculated using the formula for permutations with repetitions:

nPr = n^r

where n is the number of options and r is the number of places. In this case, n = 4 (as we have four digits) and r = 4 (as we are forming 4-digit numbers). Therefore:

nPr = 4^4 = 256

2. Identifying the pattern:
To find the sum of all 4-digit numbers, we need to identify the pattern in the digits' positions. As the digits 2, 4, 6, and 8 are all even, we can observe that each digit will occur an equal number of times in each place value.

3. Calculation:
Let's consider a single digit first. Since each digit will occur an equal number of times in each place value, we can calculate the sum of all possible combinations for a single digit and then multiply it by the total number of digits.

The sum of all possible combinations for a single digit can be found by summing the digit with itself three times (as there are four positions). Therefore, the sum for a single digit is given by:

digit_sum = (digit + digit + digit + digit)

In this case, digit_sum = (2 + 2 + 2 + 2) = 8

Now, we multiply the digit_sum by the total number of digits (4) to get the sum of all possible combinations for a single digit:

total_sum_single_digit = digit_sum * total_number_of_digits
= 8 * 4
= 32

Finally, we multiply the total_sum_single_digit by the total number of permutations to get the sum of all 4-digit numbers:

sum_of_all_numbers = total_sum_single_digit * total_number_of_permutations
= 32 * 256
= 8192

Therefore, the sum of all 4-digit numbers that can be formed using the digits 2, 4, 6, and 8 with repetitions allowed is 8192.

Formula: n^(n-1) * ( sum of n digits) * (1111...of n times)
n= 4
64*(2+4+6+8)(1111)
answer is 1422080