NavGanitMateby NavGreen Compute, AustraliaBased on Australian Curriculum: Mathematics v9.0
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Prime factors and powers

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A library in Melbourne has 25 identical shelves, each holding 31 identical books. The librarian wants to reorganise the books into boxes, with each box containing the same number of books and using the smallest possible number of boxes.

What is the smallest number of boxes needed if each box must contain a number of books that is a power of a single prime number? Express your answer as a single exact integer.

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Prime factor tree for twenty-four. 24 splits into 2 × 12; 12 into 2 × 6; 6 into 2 × 3. The prime leaves give 24 = 2³ × 3.24212262324 = 2³ × 3
24 splits into 2 × 12; 12 into 2 × 6; 6 into 2 × 3. The prime leaves give 24 = 2³ × 3.
Four squared. 4² = 4 × 4 = 16 unit squares.4 units across4
4² = 4 × 4 = 16 unit squares.

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Explanation

Worked solution and answer (click to reveal)

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Worked solution

  1. The library has 25 shelves, each with 31 books. Total books = 25 × 31.
  2. Calculate total books: 25 × 31 = 775 books.
  3. The librarian wants to pack all 775 books into boxes, each containing a number of books that is a power of a single prime number (e.g., 2^3 = 8, 3^2 = 9, 5^1 = 5).
  4. To use the smallest number of boxes, each box must hold the largest possible number of books that is a prime power and divides 775 exactly.
  5. Factor 775 into prime factors. Start by dividing by 5: 775 ÷ 5 = 155. 155 ÷ 5 = 31. 31 is a prime number.
  6. So, 775 = 5 × 5 × 31 = 5^2 × 31^1.
  7. The prime factorisation is 5^2 × 31^1. The possible prime powers that divide 775 are: 5^1, 5^2, 31^1.
  8. List all valid prime power divisors: 5, 25, 31.
  9. To minimise the number of boxes, choose the largest valid prime power divisor: 31.
  10. Number of boxes = total books ÷ books per box = 775 ÷ 31 = 25.
  11. Check: 25 boxes × 31 books = 775 books. Correct.
  12. The smallest number of boxes is 25, since 31 is a prime power and divides 775 exactly.

Reference answer: 25

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