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Question

Difficulty: MediumFactors, Multiples, and Prime Factorization

An architect is planning to pave a rectangular courtyard using exactly 240240 identical square stones. To ensure the space is functional, both the length and the width of the courtyard must consist of at least 55 stones. Assuming the stones cannot be cut, how many different unordered pairs of dimensions (length and width) are possible for this courtyard?

  1. A
    16
  2. B
    10
  3. 6Answer
  4. D
    20

Answer

6
The correct answer is derived by finding the total number of factor pairs for 240 (which is 10 pairs) and subtracting the 4 pairs that contain a factor smaller than 5 (1, 2, 3, and 4). This leaves exactly 6 valid unordered pairs.

Step-by-Step Solution

1
Find the prime factorization of 240.
240=24×31×51240 = 2^4 \times 3^1 \times 5^1
Prime factorization is required to calculate the total number of factors (divisors) of the number.
2
Calculate the total number of factors for 240.
Total factors = (4+1)(1+1)(1+1)=5×2×2=20(4+1)(1+1)(1+1) = 5 \times 2 \times 2 = 20.
Adding 1 to each exponent in the prime factorization and multiplying them yields the total number of divisors.
3
Calculate the total number of unordered pairs of dimensions.
Total pairs = 20÷2=1020 \div 2 = 10.
Since 240 is not a perfect square, its 20 factors form exactly 10 distinct pairs (a,b)(a, b) such that a×b=240a \times b = 240.
4
Identify and exclude pairs where at least one dimension is less than 5.
The factors less than 5 are 1, 2, 3, and 4. These correspond to 4 invalid pairs: (1, 240), (2, 120), (3, 80), and (4, 60).
The problem specifically states that both dimensions must consist of at least 5 stones.
5
Calculate the final number of valid pairs.
10−4=610 - 4 = 6 valid pairs.
Subtracting the 4 invalid pairs from the total 10 pairs leaves the number of pairs that satisfy all conditions.

Key Concept

Factors, Multiples, and Prime Factorization

Alternative Method

Instead of finding all factors using prime factorization, one could systematically list the factor pairs starting from 1×2401 \times 240 up to the square root of 240240 (which is approximately 15.415.4). The pairs are (1,240), (2,120), (3,80), (4,60), (5,48), (6,40), (8,30), (10,24), (12,20), and (15,16). Eliminating the first four pairs that contain a number smaller than 5 leaves exactly 6 valid options.
Estimated Time:1m 30s
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Topics

Factors, Multiples, and Prime Factorization

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