Key Concepts and Formulas

• Complementary Counting: The number of ways to have at least one of something is the total number of possibilities minus the number of ways to have none of that thing.
• Fundamental Principle of Counting: If there are $m$ ways to do one thing and $n$ ways to do another, there are $m \times n$ ways to do both.
• Sum Rule of Counting: If there are $m$ ways to do one thing and $n$ ways to do another, and the two things cannot be done simultaneously, then there are $m+n$ ways to do either one.

Step-by-Step Solution

Step 1: Define the Character Set

We have a set of 10 characters to choose from: 5 letters (A, B, C, D, E) and 5 numbers (1, 2, 3, 4, 5).

Step 2: Calculate the total number of passwords of length 6, 7, and 8

• For a password of length 6, each character can be any of the 10 characters. So, the total number of possible passwords is $10^6$.
• For a password of length 7, the total number of possible passwords is $10^7$.
• For a password of length 8, the total number of possible passwords is $10^8$.

Step 3: Calculate the number of passwords of length 6, 7, and 8 containing ONLY letters

• For a password of length 6, if it contains only letters, each character can be any of the 5 letters. So, the total number of such passwords is $5^6$.
• For a password of length 7, the total number of such passwords is $5^7$.
• For a password of length 8, the total number of such passwords is $5^8$.

Step 4: Calculate the number of passwords of length 6, 7, and 8 with at least one number

Using complementary counting, the number of passwords with at least one number is the total number of passwords minus the number of passwords with only letters:

• Length 6: $10^6 - 5^6$
• Length 7: $10^7 - 5^7$
• Length 8: $10^8 - 5^8$

Step 5: Calculate the total number of passwords with at least one number

The total number of passwords in $S$ with at least one number is the sum of the number of such passwords of length 6, 7, and 8: $(10^6 - 5^6) + (10^7 - 5^7) + (10^8 - 5^8)$ $= (10^6 + 10^7 + 10^8) - (5^6 + 5^7 + 5^8)$

Step 6: Factor out $5^6$

$= 5^6 (2^6 + 2^7 \cdot 5 + 2^8 \cdot 5^2) - 5^6(1 + 5 + 5^2)$ $= 5^6 (2^6 + 5 \cdot 2^7 + 25 \cdot 2^8 - 1 - 5 - 25)$ $= 5^6 (64 + 5 \cdot 128 + 25 \cdot 256 - 31)$ $= 5^6 (64 + 640 + 6400 - 31)$ $= 5^6 (7104 - 31)$ $= 5^6 (7073)$

Step 7: Find the value of $\alpha$

We are given that the number of passwords is $\alpha \times 5^6$. Comparing this with our result, we have: $\alpha \times 5^6 = 7073 \times 5^6$ Therefore, $\alpha = 7073$.

Common Mistakes & Tips

• Forgetting Complementary Counting: It's easy to try to directly count the number of passwords with at least one number, but this is much more complicated than using complementary counting.
• Miscalculating Powers: Be careful when calculating the powers of 5 and 10.
• Algebraic Errors: Ensure that all algebraic manipulations are correct, especially when factoring and simplifying expressions.

Summary

We found the total number of passwords with at least one number by first finding the total number of passwords of lengths 6, 7, and 8, and then subtracting the number of passwords of those lengths consisting only of letters. This gave us an expression of the form $\alpha \times 5^6$, and we were able to solve for $\alpha$.

The final answer is \boxed{7073}.