Let's solve the problem based on the details provided in the image:

Method A:

• First digit: It must be an odd number between 0 and 10. The odd numbers in this range are: 1, 3, 5, 7, 9. This gives 5 possibilities.
• Last two digits: Both must be multiples of 10. In the range from 0 to 9, the only multiple of 10 is 0. This gives only 1 possibility for each of the last two digits.

Thus, the total number of codes for Method A is: $5 \times 1 \times 1 = 5 \text{ codes}$

Method B:

• First digit: It must be an odd number (but not 0). The odd numbers are: 1, 3, 5, 7, 9. This gives 5 possibilities.
• Second digit: It must be an even number (but not 0). The even numbers are: 2, 4, 6, 8. This gives 4 possibilities.
• Third digit: It must be an odd number (but not 0). The odd numbers are again: 1, 3, 5, 7, 9. This gives 5 possibilities.

Thus, the total number of codes for Method B is: $5 \times 4 \times 5 = 100 \text{ codes}$

Difference:

To find how many more codes can be made with Method B than with Method A, subtract the number of Method A codes from the number of Method B codes: $100 - 5 = 95$

Thus, Method B can make 95 more codes than Method A.

The answer is:
$\boxed{95}$

Would you like any further clarifications or detailed steps?

Here are 5 relative questions to explore further:

1. How would the solution change if repetition of digits was not allowed?
2. What if Method A allowed any digit for the last two positions instead of just multiples of 10?
3. How many codes can be made if zero was allowed for the second digit in Method B?
4. How does the solution change if we use 4-digit codes instead of 3-digit codes?
5. Can you create your own method for generating 3-digit codes and calculate how many possible combinations exist?

Tip: Always break down problems like this into simple, logical steps, and list all possibilities to avoid missing any cases!