Math Problem Statement
Solution
The problem asks us to find how many books the librarian has based on the following conditions:
- If the librarian puts 2, 3, 4, 5, or 6 books on each shelf, there is always 1 book left on the last shelf.
- If the librarian puts 7 books on each shelf, the books fit exactly.
We need to find the smallest number of books that satisfies these conditions.
Step-by-step explanation:
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Condition 1: For 2, 3, 4, 5, and 6 books per shelf, there is always 1 book left. This implies: Therefore, 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, must be a multiple of 60, meaning:
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Condition 2: If the librarian puts 7 books on each shelf, the books fit exactly. This implies: Combining this with the equation from Condition 1, we have: Simplifying: Since , we now have: To solve , we multiply both sides by the modular inverse of 4 modulo 7. The inverse of 4 modulo 7 is 2 (because ). Multiplying both sides by 2: Thus, for some integer .
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Final expression for : Substituting into , we get: Therefore, .
Smallest solution:
For , the smallest value of is:
Thus, the librarian has 301 new books.
Would you like more details or have any questions? Here are some related questions you can explore:
- What is the general approach for solving problems involving the Chinese Remainder Theorem?
- How is the least common multiple (LCM) used in modular arithmetic problems?
- What is the modular inverse, and how is it calculated?
- How can this type of problem be generalized to other divisors and remainders?
- 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.
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Math Problem Analysis
Mathematical Concepts
Modular Arithmetic
Least Common Multiple (LCM)
Chinese Remainder Theorem
Formulas
N ≡ 1 (mod 2)
N ≡ 1 (mod 3)
N ≡ 1 (mod 4)
N ≡ 1 (mod 5)
N ≡ 1 (mod 6)
N ≡ 0 (mod 7)
Theorems
Chinese Remainder Theorem
Suitable Grade Level
Grade 10-12
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