Zizi had some cards. She could place them into sets of 4 cards or sets of 6 cards
without leaving any remainder. When she placed them into sets of 7 cards instead,
she needed one more card to make the last set of 7. What was the smallest
possible number of cards she had?

To determine the smallest possible number of cards Zizi had, let's denote the number of cards as \( N \).

We know the following:
1. \( N \) is divisible by 4.
2. \( N \) is divisible by 6.
3. \( N \) leaves a remainder of 1 when divided by 7.

From the first two conditions, we can conclude that \( N \) must be a common multiple of 4 and 6. The least common multiple (LCM) of 4 and 6 is:

\[
\text{LCM}(4, 6) = 12
\]

Thus, \( N \) must be a multiple of 12. We can express \( N \) as:

\[
N = 12k
\]

for some integer \( k \).

Now, considering the third condition:

\[
N \equiv 1 \pmod{7}
\]

Substituting \( N = 12k \) into the congruence gives:

\[
12k \equiv 1 \pmod{7}
\]

We need to find the smallest positive integer \( k \) such that this congruence holds. To solve this, we can simplify \( 12 \mod 7 \):

\[
12 \equiv 5 \pmod{7}
\]

Thus, the congruence becomes:

\[
5k \equiv 1 \pmod{7}
\]

We need to find the multiplicative inverse of 5 modulo 7, which is an integer \( k \) such that:

\[
5k \equiv 1 \pmod{7}
\]

By testing values, we find that:

\[
5 \times 3 = 15 \equiv 1 \pmod{7}
\]

Thus, \( k = 3 \) is the multiplicative inverse. Therefore, the smallest \( k \) is 3. Substituting \( k = 3 \) back into \( N = 12k \):

\[
N = 12 \times 3 = 36
\]

To verify:
- 36 is divisible by 4: \( 36 \div 4 = 9 \)
- 36 is divisible by 6: \( 36 \div 6 = 6 \)
- 36 leaves a remainder of 1 when divided by 7: \( 36 \div 7 = 5 \) remainder 1

Therefore, the smallest possible number of cards Zizi had is:

\[
\boxed{36}
\]

Would you like any more details or have any questions?

Here are some questions for you to consider:
1. What is the least common multiple (LCM) of two numbers?
2. How do you find the multiplicative inverse in modular arithmetic?
3. What is the process to verify if a number satisfies given conditions in a problem?
4. How can congruences be used to solve real-world problems?
5. What are some other methods to solve modular arithmetic problems?

**Tip:** When dealing with divisibility and remainders, modular arithmetic is a powerful tool that simplifies complex problems.

Discover how to find the smallest possible number of cards Zizi had using modular arithmetic. Solve the problem involving divisibility by 4 and 6, and a remainder condition with 7. Detailed step-by-step solution provided.

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