The addition problem above shows four of the 24 different integers tha

To find the sum of all 24 integers formed by using each of the digits 1, 2, 3, and 4 exactly once, we can use a bit of logic. Since each digit will appear in each position (units, tens, hundreds, and thousands) equally, we can start by finding the sum of all the possible permutations of the digits.

To do this, we can use the formula:

n!

where n is the total number of items (in this case, 4), and r is the number of items we are selecting at a time (also 4, since we are using all four digits in each number).

So the total number of possible permutations of the digits is:

4! = 24

Each of these permutations can be used as the units digit in one of the 24 numbers in the sum. So to find the sum of all 24 numbers, we simply need to add up the permutations in each of the other three positions (tens, hundreds, and thousands).

Since each digit appears in each position equally, the sum of all the permutations in each position will be the same. So we can find the sum of all the permutations in each position by adding up the digits (1+2+3+4=10) and multiplying by the number of permutations:

10 * 6 = 60

So the sum of all 24 integers formed by using each of the digits 1, 2, 3, and 4 exactly once is:

60 + 600 + 6000 + 60000 = 66660