Using each of the digits 1, 3, 5 and 7 exactly once in each number, how many 4 digit numbers can be formed in total, and what is the sum of all such 4 digit numbers?

Introduction / Context:
This problem combines permutations with place value concepts. We are asked to form all possible 4 digit numbers using the digits 1, 3, 5 and 7, with each digit used exactly once in each number, and then to compute the sum of all such numbers.

Given Data / Assumptions:

• Digits available: 1, 3, 5, 7.
• Each 4 digit number must use each digit exactly once.
• No repetition of digits within a single number.
• All permutations of these four digits are valid 4 digit numbers.

Concept / Approach:
The number of distinct 4 digit numbers formed from these four digits is 4!, since all digits are distinct. Each digit appears an equal number of times in each place value (units, tens, hundreds, thousands). We can exploit this symmetry to compute the total sum without listing every number.

Step-by-Step Solution:
Step 1: Count the total number of permutations: 4! = 4 * 3 * 2 * 1 = 24. Step 2: Each of the four digits appears equally often in each place. In 24 permutations, each digit appears 24 / 4 = 6 times in any given position. Step 3: Sum of the digits = 1 + 3 + 5 + 7 = 16. Step 4: Contribution of the thousands place = 6 * 16 * 1000. Step 5: Contribution of the hundreds place = 6 * 16 * 100. Step 6: Contribution of the tens place = 6 * 16 * 10. Step 7: Contribution of the units place = 6 * 16 * 1. Step 8: Factor out 6 * 16 = 96. Total sum = 96 * (1000 + 100 + 10 + 1) = 96 * 1111. Step 9: Compute 96 * 1111 = 106656.

Verification / Alternative check:
As a check, you could compute the sum for a smaller example with two or three digits and verify the pattern. The symmetry argument that each digit appears the same number of times in each position remains valid and scales correctly to this case as well.

Why Other Options Are Wrong:
105555, 106665 and 108333 are near but not equal to the computed total. They reflect small arithmetic mistakes or incorrectly grouped contributions from place values.

Common Pitfalls:
Mistakes often occur in counting how many times each digit appears in each place or in adding up place value contributions. Some students attempt to list all 24 numbers manually, which is time consuming and prone to error.

Final Answer:
The sum of all 4 digit numbers formed using 1, 3, 5 and 7 exactly once in each number is 106656.