Question

1) The smallest perfect square number 1 pointdivisible by 5, 6 and 27 is​

Answer 1

Answer:

8100 is the smallest perfect square divisible by 5,6 and 27

Step-by-step explanation:

5 = 5 * 1

6 = 2 * 3

27 = 3 * 3 *3

5 * 6 * 27 = 2 * 3 * 3 * 3 * 3 * 5

Factors of perfect square will be perfect squares

To make this a perfect, multiply by 5 * 2

Perfect square = 5 * 6 * 27 * 5 * 2

                        = 8100

Answer 2

Answer:

$8100$, which is the square of $90$.

Step-by-step explanation:

Factor the three divisors into prime numbers:

• $5$ is a prime number itself.
• $6 = 2\times 3$.
• $27 = 3^3$.

Any number divisible by these three divisors should include the following factors:

• $2$,
• $3^3$, and
• $5$.

Note that all these three prime factors have an odd power ($1$, $3$, and $1$, respectively)

• $2$ becomes $2^2$ (add $1$ to the initial power of $1$ to obtain $2$.)
• $3^3$ becomes $3^4$.
• $5$ becomes $5^2$.

The product of these three factors would be:

$2^2 \times 3^4 \times 5^2 = 8100$.

Indeed, $8100$ is divisible by all these three divisors. At the same time, because all the powers of its prime factors are even,

$8100 = (2 \times 3^2 \times 5)^2 = 90^2$.