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can you pleas do it with 6th grade math and can i see your calculations 

Let's solve this problem step by step using concepts suitable for a 6th-grade level.

### Problem Recap:
The librarian is putting new books on shelves:
- If she puts 2, 3, 4, 5, or 6 books per shelf, there is always 1 book left over.
- If she puts 7 books per shelf, there are no books left over.

We need to find out how many books the librarian has.

### Step-by-step Solution:

1. **Understanding the Remainders**:
   - The number of books \( N \) leaves a remainder of 1 when divided by 2, 3, 4, 5, or 6.
   - This means \( N \) is 1 more than a common multiple of these numbers.

2. **Finding the Least Common Multiple (LCM)**:
   - Find the LCM of 2, 3, 4, 5, and 6:
     - \( \text{LCM of 2 and 3} = 6 \)
     - \( \text{LCM of 6 and 4} = 12 \)
     - \( \text{LCM of 12 and 5} = 60 \)
     - \( \text{LCM of 60 and 6} = 60 \)
   - So, the LCM of 2, 3, 4, 5, and 6 is 60.

3. **Forming the Equation**:
   - Since \( N \) leaves a remainder of 1 when divided by 2, 3, 4, 5, or 6, we can express this as:
     \[
     N = 60k + 1
     \]
     where \( k \) is an integer.

4. **Considering Division by 7**:
   - We also know that \( N \) is divisible by 7. This means:
     \[
     60k + 1 \mod 7 = 0
     \]
   - We need to find a value of \( k \) such that \( N = 60k + 1 \) is divisible by 7.

5. **Testing Values of \( k \)**:
   - Try \( k = 1 \):
     \[
     N = 60 \times 1 + 1 = 61 \quad \text{(not divisible by 7)}
     \]
   - Try \( k = 2 \):
     \[
     N = 60 \times 2 + 1 = 121 \quad \text{(not divisible by 7)}
     \]
   - Try \( k = 3 \):
     \[
     N = 60 \times 3 + 1 = 181 \quad \text{(not divisible by 7)}
     \]
   - Try \( k = 4 \):
     \[
     N = 60 \times 4 + 1 = 241 \quad \text{(not divisible by 7)}
     \]
   - Try \( k = 5 \):
     \[
     N = 60 \times 5 + 1 = 301 \quad \text{(divisible by 7)}
     \]
     Since \( 301 \div 7 = 43 \) (a whole number), \( N = 301 \) satisfies all conditions.

### Final Answer:
The librarian has **301 books**.

Would you like further clarification on any of these steps, or have any questions?

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Here are 5 related questions that might deepen your understanding of this topic:

1. What is the importance of finding the least common multiple in problems like this?
2. How can we check if a number is divisible by another number without performing full division?
3. What is the difference between factors and multiples?
4. How would the solution change if the number left over was 2 instead of 1?
5. How could you use a calculator to check your answers more quickly?

**Tip**: When solving word problems, breaking down the conditions one by one can make it easier to translate them into equations or inequalities!

The librarian is putting 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?

In this problem, a librarian is shelving books. Learn how to solve using least common multiples and division rules to find how many books fit exactly into the given shelf arrangements. Suitable for Grade 6 students.

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