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The problem asks us to find how many books the librarian has based on the following conditions:

1. If the librarian puts 2, 3, 4, 5, or 6 books on each shelf, there is always **1 book left** on the last shelf.
2. If the librarian puts 7 books on each shelf, the books fit **exactly**.

We need to find the smallest number \( N \) of books that satisfies these conditions.

### Step-by-step explanation:

1. **Condition 1:**
   For 2, 3, 4, 5, and 6 books per shelf, there is always 1 book left. This implies:
   \[
   N \equiv 1 \pmod{2}, \quad N \equiv 1 \pmod{3}, \quad N \equiv 1 \pmod{4}, \quad N \equiv 1 \pmod{5}, \quad N \equiv 1 \pmod{6}
   \]
   Therefore, \( N - 1 \) must be divisible by 2, 3, 4, 5, and 6. The least common multiple (LCM) of 2, 3, 4, 5, and 6 is 60. So, \( N - 1 \) must be a multiple of 60, meaning:
   \[
   N = 60k + 1 \quad \text{for some integer } k.
   \]

2. **Condition 2:**
   If the librarian puts 7 books on each shelf, the books fit exactly. This implies:
   \[
   N \equiv 0 \pmod{7}.
   \]
   Combining this with the equation from Condition 1, we have:
   \[
   60k + 1 \equiv 0 \pmod{7}.
   \]
   Simplifying:
   \[
   60k + 1 \equiv 0 \pmod{7} \quad \Rightarrow \quad 60k \equiv -1 \pmod{7}.
   \]
   Since \( 60 \equiv 4 \pmod{7} \), we now have:
   \[
   4k \equiv -1 \pmod{7}.
   \]
   To solve \( 4k \equiv -1 \pmod{7} \), we multiply both sides by the modular inverse of 4 modulo 7. The inverse of 4 modulo 7 is 2 (because \( 4 \times 2 = 8 \equiv 1 \pmod{7} \)). Multiplying both sides by 2:
   \[
   k \equiv -2 \pmod{7} \quad \Rightarrow \quad k \equiv 5 \pmod{7}.
   \]
   Thus, \( k = 7m + 5 \) for some integer \( m \).

3. **Final expression for \( N \):**
   Substituting \( k = 7m + 5 \) into \( N = 60k + 1 \), we get:
   \[
   N = 60(7m + 5) + 1 = 420m + 301.
   \]
   Therefore, \( N = 420m + 301 \).

### Smallest solution:
For \( m = 0 \), the smallest value of \( N \) is:
\[
N = 301.
\]

Thus, the librarian has **301 new books**.

---

Would you like more details or have any questions? Here are some related questions you can explore:

1. What is the general approach for solving problems involving the Chinese Remainder Theorem?
2. How is the least common multiple (LCM) used in modular arithmetic problems?
3. What is the modular inverse, and how is it calculated?
4. How can this type of problem be generalized to other divisors and remainders?
5. What are some practical applications of this type of number theory problem?

**Tip:** When solving modular arithmetic problems, breaking them into smaller, manageable conditions and using tools like the LCM and modular inverses can help you find the solution efficiently.

The librarian is putting the new books on the shelves. If she puts 2, 3, 4, 5, or 6 books on each shelf, there is always one book on the last shelf. If she puts 7 books on each shelf, they fit exactly. How many new books does the librarian have? (Imagine the librarian has infinite shelves.)

This problem involves modular arithmetic and asks how many books a librarian has based on specific conditions when placing books on shelves. Using the least common multiple (LCM) and the Chinese Remainder Theorem, the solution shows how to determine the smallest number of books, with the result being 301 books.

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