Question

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

Answer 1

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

Explanation:

Factor the three divisors into prime numbers:

• 
  `5` is a prime number itself.
• 
  `6 = 2* 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 * 3^4 * 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 * 3^2 * 5)^2 = 90^2`.

Answer 2

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

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