Finding the Least Square Number Divisible by Multiple Numbers

The question asks for the least square number that is exactly divisible by 2, 3, 10, 18, and 20. A square number is an integer that is the square of another integer (e.g., 4, 9, 16, 25, 36, 100, 900 are square numbers).

For a number to be exactly divisible by a set of numbers, it must be a multiple of their Least Common Multiple (LCM).

Step 1: Find the LCM of 2, 3, 10, 18, and 20

We find the LCM by first determining the prime factorization of each number:

• Prime factorization of 2: \(2 = 2^1\)
• Prime factorization of 3: \(3 = 3^1\)
• Prime factorization of 10: \(10 = 2 \times 5 = 2^1 \times 5^1\)
• Prime factorization of 18: \(18 = 2 \times 9 = 2 \times 3^2 = 2^1 \times 3^2\)
• Prime factorization of 20: \(20 = 4 \times 5 = 2^2 \times 5 = 2^2 \times 5^1\)

To find the LCM, we take the highest power of each prime factor that appears in any of the factorizations:

• Highest power of 2: \(2^2\) (from 20)
• Highest power of 3: \(3^2\) (from 18)
• Highest power of 5: \(5^1\) (from 10 and 20)

LCM is the product of these highest powers:

\( \text{LCM}(2, 3, 10, 18, 20) = 2^2 \times 3^2 \times 5^1 = 4 \times 9 \times 5 = 36 \times 5 = 180 \)

The least number exactly divisible by 2, 3, 10, 18, and 20 is 180.

Step 2: Find the Least Square Number which is a Multiple of 180

The number we are looking for must be a multiple of 180 AND a perfect square. Let's look at the prime factorization of 180 again: \(180 = 2^2 \times 3^2 \times 5^1\).

For a number to be a perfect square, the exponents of all its prime factors must be even. In the factorization of 180, the exponents are 2 (for 2), 2 (for 3), and 1 (for 5). The exponent for the prime factor 5 is odd (1).

To make the number a perfect square, we need to multiply 180 by the smallest factor that will make all exponents even. We need to increase the exponent of 5 from 1 to the next even number, which is 2. This requires multiplying by \(5^{2-1} = 5^1 = 5\).

The least square number divisible by 180 is therefore:

\( 180 \times 5 = (2^2 \times 3^2 \times 5^1) \times 5^1 = 2^2 \times 3^2 \times 5^2 \)

Now, calculate the value:

\( 2^2 \times 3^2 \times 5^2 = 4 \times 9 \times 25 = 36 \times 25 = 900 \)

Step 3: Verify the Result

The number 900 is a perfect square because \(30^2 = 900\).

Let's check if 900 is divisible by 2, 3, 10, 18, and 20:

• \(900 \div 2 = 450\) (Yes)
• \(900 \div 3 = 300\) (Yes)
• \(900 \div 10 = 90\) (Yes)
• \(900 \div 18 = 50\) (Yes)
• \(900 \div 20 = 45\) (Yes)

Since 900 is a square number and is divisible by all the given numbers, and it was constructed as the least multiple of the LCM that is a square, it is the least square number with this property.

Comparing with the Options

• 30: Not a square number. Not divisible by 18 or 20.
• 180: Not a square number (\(13^2 = 169\), \(14^2 = 196\)). It is divisible by all given numbers, but not a square.
• 196: A square number (\(14^2 = 196\)). Not divisible by 3, 10, 18, or 20.
• 900: A square number (\(30^2 = 900\)). Divisible by 2, 3, 10, 18, and 20. This matches our calculated value.

Therefore, 900 is the least square number exactly divisible by 2, 3, 10, 18, and 20.

Revision Table: Key Concepts

Additional Information: Divisibility and Perfect Squares

Understanding divisibility rules and the properties of perfect squares is crucial for solving such problems.

• Divisibility: If a number 'A' is divisible by numbers \(n_1, n_2, ..., n_k\), then 'A' must be a multiple of the LCM of \((n_1, n_2, ..., n_k)\).
• Perfect Squares: A number N is a perfect square if and only if in its prime factorization, \(N = p_1^{a_1} p_2^{a_2} ... p_k^{a_k}\), all the exponents \(a_1, a_2, ..., a_k\) are even integers (\(0, 2, 4, ...\)).
• To make any positive integer N a perfect square by multiplying it by the smallest possible integer, first find the prime factorization of N. Identify the prime factors that have odd exponents. For each such prime factor \(p^a\) (where 'a' is odd), you need to multiply by \(p^1\) to make the exponent even (\(p^{a+1}\) where \(a+1\) is even). The smallest multiplier is the product of all such prime factors raised to the power of 1.

In this problem, the LCM was \(180 = 2^2 \times 3^2 \times 5^1\). The factor \(5^1\) has an odd exponent (1). To make it a perfect square, we multiplied by \(5^1\), resulting in \(2^2 \times 3^2 \times 5^2 = 900\).