Math Problem Statement

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.)

Solution

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 NN 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: N1(mod2),N1(mod3),N1(mod4),N1(mod5),N1(mod6)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, N1N - 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, N1N - 1 must be a multiple of 60, meaning: N=60k+1for some integer k.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: N0(mod7).N \equiv 0 \pmod{7}. Combining this with the equation from Condition 1, we have: 60k+10(mod7).60k + 1 \equiv 0 \pmod{7}. Simplifying: 60k+10(mod7)60k1(mod7).60k + 1 \equiv 0 \pmod{7} \quad \Rightarrow \quad 60k \equiv -1 \pmod{7}. Since 604(mod7)60 \equiv 4 \pmod{7}, we now have: 4k1(mod7).4k \equiv -1 \pmod{7}. To solve 4k1(mod7)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 4×2=81(mod7)4 \times 2 = 8 \equiv 1 \pmod{7}). Multiplying both sides by 2: k2(mod7)k5(mod7).k \equiv -2 \pmod{7} \quad \Rightarrow \quad k \equiv 5 \pmod{7}. Thus, k=7m+5k = 7m + 5 for some integer mm.

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

Smallest solution:

For m=0m = 0, the smallest value of NN is: N=301.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.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

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