Question

What is the smallest square number which is divisible by 4, 6 and 32?

• A. \( 100 \)
• B. \( 196 \)
• C. \( 96 \)
• D. \( 576 \)

Hint:

\textbf{Smallest Square Divisible by Numbers.} To find the smallest square number divisible by a set of numbers, first find the LCM of the numbers. Then, find the prime factorization of the LCM. For each prime factor with an odd exponent, multiply the LCM by that prime factor to make the exponent even. The resulting number will be the smallest square number divisible by the original set of numbers.

Answer 1

The Correct Option is D

We need to find the smallest square number that is divisible by 4, 6, and 3(B) First, find the least common multiple (LCM) of 4, 6, and 3(B) Prime factorization of the numbers: \( 4 = 2^2 \) \( 6 = 2 \times 3 \) \( 32 = 2^5 \) The LCM is the product of the highest powers of all prime factors involved: LCM(4, 6, 32) \( = 2^5 \times 3^1 = 32 \times 3 = 96 \) Now, we need to find the smallest square number that is a multiple of 96. Prime factorization of 96 is \( 2^5 \times 3^1 \). For a number to be a perfect square, all the exponents in its prime factorization must be even. In the prime factorization of 96, the exponent of 2 is 5 (odd) and the exponent of 3 is 1 (odd). To make this number a perfect square, we need to multiply it by the smallest number that will make all the exponents even. We need to multiply by \( 2^{6-5} = 2^1 \) and \( 3^{2-1} = 3^1 \). So, the required multiplier is \( 2 \times 3 = 6 \). The smallest square number divisible by 4, 6, and 32 is \( 96 \times 6 \): \( 96 \times 6 = (2^5 \times 3^1) \times (2^1 \times 3^1) = 2^{5+1} \times 3^{1+1} = 2^6 \times 3^2 = 64 \times 9 = 576 \) We can check if 576 is a perfect square: \( \sqrt{576} = 24 \). We can also check if 576 is divisible by 4, 6, and 32: \( 576 \div 4 = 144 \) \( 576 \div 6 = 96 \) \( 576 \div 32 = 18 \) Therefore, the smallest square number which is divisible by 4, 6, and 32 is 576.