Finding the Least Perfect Square Divisible by 4, 10, and 12

The question asks for the least number that satisfies two conditions:

1. It must be a perfect square.
2. It must be divisible by each of the numbers 4, 10, and 12.

Understanding Divisibility and LCM

If a number is divisible by 4, 10, and 12, it must be a common multiple of these numbers. To find the least such number that is a multiple of all three, we need to calculate the Least Common Multiple (LCM) of 4, 10, and 12.

We find the LCM using prime factorization:

• Prime factorization of 4: $4 = 2 \times 2 = 2^2$
• Prime factorization of 10: $10 = 2 \times 5 = 2^1 \times 5^1$
• Prime factorization of 12: $12 = 2 \times 2 \times 3 = 2^2 \times 3^1$

The LCM is found by taking the highest power of each prime factor that appears in any of the factorizations:

LCM(4, 10, 12) = $2^{\max(2,1,2)} \times 3^{\max(0,0,1)} \times 5^{\max(0,1,0)}$

LCM(4, 10, 12) = $2^2 \times 3^1 \times 5^1 = 4 \times 3 \times 5 = 60$

So, any number divisible by 4, 10, and 12 must be a multiple of 60. The required number must be of the form $60k$ for some integer $k$.

Understanding Perfect Squares

A perfect square is an integer that is the square of another integer. In terms of prime factorization, a number is a perfect square if and only if all the exponents in its prime factorization are even.

Finding the Least Perfect Square Multiple of 60

The number we are looking for is a multiple of 60 and a perfect square. Let the number be $N$.

$N = 60k$

We know the prime factorization of 60 is $2^2 \times 3^1 \times 5^1$.

So, $N = (2^2 \times 3^1 \times 5^1) \times k$

For $N$ to be a perfect square, all exponents in its prime factorization must be even. In the factorization of 60, the exponents of 2, 3, and 5 are 2, 1, and 1 respectively. The exponents of 3 and 5 are odd.

To make the exponents of 3 and 5 even, the smallest value of $k$ must contribute at least $3^1$ and $5^1$. The exponent of 2 is already even (2), so $k$ doesn't need to contribute any factors of 2 to make it even. Any other prime factors in $k$ would also need to have even exponents.

The least value for $k$ that makes $N$ a perfect square is $k = 3^1 \times 5^1 = 15$.

Now, we calculate the least perfect square number:

$N = 60 \times k = 60 \times 15$

$N = (2^2 \times 3^1 \times 5^1) \times (3^1 \times 5^1)$

$N = 2^{2+0} \times 3^{1+1} \times 5^{1+1}$

$N = 2^2 \times 3^2 \times 5^2$

Since all exponents (2, 2, 2) are even, $N$ is a perfect square.

$N = (2 \times 3 \times 5)^2 = 30^2 = 900$

The least number which is a perfect square and is divisible by 4, 10, and 12 is 900.

Verifying with Options

Let's quickly check the given options:

From the table, only 900 is a perfect square divisible by 4, 10, and 12. Since our calculation also yielded 900 and we specifically sought the *least* number by using the LCM and minimum required factors to make it a perfect square, 900 is indeed the correct answer.

Revision Table: Least Perfect Square

Additional Information: Properties of Numbers

Understanding the properties of numbers like divisibility and perfect squares is fundamental in number theory. Here are some key points:

• Divisibility Rules: Knowing divisibility rules (e.g., a number is divisible by 4 if its last two digits are divisible by 4; by 10 if it ends in 0; by 12 if it's divisible by both 3 and 4) can help quickly check options. For 900: ends in 00 (divisible by 4), ends in 0 (divisible by 10), sum of digits $9+0+0=9$ (divisible by 3) and ends in 00 (divisible by 4), so divisible by 12.
• LCM and GCF: LCM is useful for problems involving events repeating at intervals or finding a common point (like shared divisibility). The Greatest Common Factor (GCF) is useful for problems involving division into equal parts.
• Perfect Square Recognition: Besides checking exponents in prime factorization, you can recognize perfect squares by their square roots being integers. Also, the number of divisors of a perfect square is always odd.
• Combining Conditions: When a problem requires a number to satisfy multiple conditions (like being a multiple of several numbers AND having another property), find the numbers that satisfy each condition separately and then find the intersection of those sets. In this case, finding multiples of LCM and finding perfect squares, then finding the smallest number common to both sets.