Each Sunday, the hymns to be sung at church are displayed on a hymn-board, using numbered
cards. The cards have the digits 0 to 9 on them. The hymn book has 750 hymns in it. Three
different hymns are sung at each service.
a. What is the smallest number of cards the church must have to be able to display any
combination of three hymns?
b. If the number 6 can be used upside down for a 9, how many cards can be saved from
your previous answer?

Let's break down the problem and solve each part step by step.

### Part (a): Minimum Number of Cards Required

1. **Understanding the Problem:**
   - There are 750 hymns, so the hymn numbers range from 1 to 750.
   - Three hymns are displayed at each service.
   - The digits 0 to 9 are used on the cards to form these hymn numbers.

2. **Determining the Number of Digits:**
   - Since there are 750 hymns, the maximum hymn number is 750, which is a three-digit number.
   - Therefore, to display any hymn number, we need cards for the digits 0 to 7. Specifically:
     - The first digit (hundreds place) can be 0, 1, 2, 3, 4, 5, 6, or 7.
     - The second digit (tens place) can be any digit from 0 to 9.
     - The third digit (units place) can be any digit from 0 to 9.

3. **Calculating the Minimum Number of Cards:**
   - We need to consider all the digits required for each place in the hymn numbers.
   - For the hundreds place: digits 0 to 7 are required (8 cards).
   - For the tens place: digits 0 to 9 are required (10 cards).
   - For the units place: digits 0 to 9 are required (10 cards).

   Since each digit could be repeated in different hymn numbers, we need to ensure there are enough cards for each digit.

4. **Counting the Maximum Number of Each Digit Required:**
   - In the worst case, all three hymns could share the same digit in the same place (e.g., hymns 111, 211, and 311 require three "1"s in the hundreds place).
   - Therefore, for each digit, we need at least three copies.

5. **Total Cards Needed:**
   - For digits 0 to 7: 8 digits × 3 copies = 24 cards.
   - For digits 0 to 9: 10 digits × 3 copies = 30 cards.
   - Since digits 0 to 7 are already counted, we need additional cards for digits 8 and 9: 2 digits × 3 copies = 6 cards.

   **Total number of cards required** = 24 (for digits 0 to 7) + 30 (for digits 0 to 9, minus overlap) + 6 (additional cards for digits 8 and 9) = 30 cards.

### Part (b): Number of Cards Saved if 6 can be Used as 9

1. **Impact of Using 6 as 9:**
   - If 6 can be used upside down as 9, then we don't need separate cards for the digit 9.
   - This means we can reduce the total number of cards by eliminating the need for extra cards specifically for the digit 9.

2. **Calculating the Savings:**
   - We originally needed 3 cards for the digit 9.
   - By using the 6 upside down, we save these 3 cards.

   **Total number of cards saved** = 3 cards.

   **New total number of cards required** = 30 cards - 3 cards = 27 cards.

### Final Answers:
a. The smallest number of cards required to display any combination of three hymns is **30**.
b. If the number 6 can be used upside down for a 9, the church can save **3** cards, reducing the total to **27** cards.

Would you like further details or have any other questions?

Here are some additional questions related to this topic:

1. How would the required number of cards change if there were four hymns displayed instead of three?
2. What if the hymn book had 1000 hymns instead of 750?
3. How many cards would be needed if the hymn book had only 99 hymns?
4. How would the problem change if only even-numbered hymns were used?
5. What if the digits 2 and 5 could also be used interchangeably, like 6 and 9?
6. How would the solution be affected if each hymn number had to be displayed in a different color?
7. What if the hymn numbers had four digits instead of three?
8. How would the required number of cards change if duplicate digits in hymn numbers were less likely?

**Tip:** When solving combinatorial problems, always consider the worst-case scenario to ensure your solution covers all possibilities.

Discover the minimum number of cards required to display any combination of three hymns at church. This problem involves combinatorics and number theory, addressing how to efficiently use cards numbered 0 to 9 for hymns numbered 1 to 750. Suitable for advanced high school students and beyond.

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