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 solve the problem, we need to find the smallest number of cards \( N \) that meets the given conditions. Specifically:

1. \( N \) can be evenly divided by 4 and 6.
2. When \( N \) is divided by 7, the remainder is 1.

First, let's find the least common multiple (LCM) of 4 and 6, as \( N \) must be a multiple of this LCM to satisfy the first condition.

The prime factorizations are:
- \( 4 = 2^2 \)
- \( 6 = 2 \times 3 \)

The LCM is obtained by taking the highest power of each prime factor:
\[ \text{LCM}(4, 6) = 2^2 \times 3 = 12 \]

So, \( N \) must be a multiple of 12:
\[ N = 12k \]
for some integer \( k \).

Next, we use the second condition:
\[ N \equiv 1 \pmod{7} \]

Substituting \( N = 12k \) into the congruence:
\[ 12k \equiv 1 \pmod{7} \]

Simplifying \( 12 \pmod{7} \):
\[ 12 \equiv 5 \pmod{7} \]

So the equation becomes:
\[ 5k \equiv 1 \pmod{7} \]

We need to find the multiplicative inverse of 5 modulo 7, which is a number \( x \) such that:
\[ 5x \equiv 1 \pmod{7} \]

We can test values for \( x \) to find the inverse:
\[ 5 \times 1 = 5 \equiv 5 \pmod{7} \]
\[ 5 \times 2 = 10 \equiv 3 \pmod{7} \]
\[ 5 \times 3 = 15 \equiv 1 \pmod{7} \]

Therefore, the multiplicative inverse of 5 modulo 7 is 3. Hence:
\[ k \equiv 3 \pmod{7} \]
\[ k = 7m + 3 \]
for some integer \( m \).

Now, substituting \( k = 7m + 3 \) back into \( N = 12k \):
\[ N = 12(7m + 3) \]
\[ N = 84m + 36 \]

The smallest \( N \) occurs when \( m = 0 \):
\[ N = 84(0) + 36 \]
\[ N = 36 \]

Thus, the smallest possible number of cards Zizi had is \( \boxed{36} \).

Would you like details or have any questions? 

Here are some related questions you might find interesting:
1. How do you find the LCM of two numbers?
2. What is the significance of congruences in number theory?
3. Can you explain the Chinese Remainder Theorem?
4. How do you determine the multiplicative inverse modulo \( n \)?
5. What are some practical applications of modular arithmetic?

**Tip:** When solving modular arithmetic problems, finding the least common multiple (LCM) and using properties of congruences are often very helpful strategies.

Discover how to find the smallest number of cards that can be placed into sets of 4 or 6 cards without remainder and needs one more card to complete a set of 7. This detailed solution involves calculating the least common multiple (LCM), solving modular congruences, and applying number theory concepts. Ideal for students in Grades 9-12.

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