Question

Find the least perfect square that is exactly divisible by 4, 5, 6, 15, and 18.
(a) 100
(b) 256
(c) 400
(d) 900

Answer 1

Hint: First find the LCM of 4, 5, 6, 15, and 18 and keep the LCM in factors form. Then multiply the LCM by the factors which are appearing odd number of times, as for a number to be a perfect square its prime factors must be appearing even number of times when written in factorised form. After multiplying the factors find the value and answer the question.

Complete step-by-step answer:
Let us start the solution to the above question by finding the LCM of 4, 5, 6, 15, and 18. For finding the LCM, we will use the method of prime factorisation.
First, we know that 4 can be written as $2\times 2$ . 5 is itself a prime, so it cannot be further factorised. 6 is the product of 2 and 3. 15 can be written as the product of two primes 3 and 5. If we talk about 18, it can be written as the product of 3 and 6, and 6 can be further written as $3\times 2$ .
So, let us represent each number in terms of its prime factors.
$4=2\times 2$
$5=5\times 1$
$6=2\times 3$
$15=3\times 5$
$18=2\times 3\times 3$
Now for finding the LCM, we count the factors the maximum number of times they have occurred in the factorised form of the number. So, 2 and 3 occur a maximum of 2 times in the factorised form of 4 and 18, respectively, while 5 occurs a maximum of a single time.
$LCM=2\times 2\times 3\times 3\times 5$
Now if we look at the LCM, 2 and 3 are occurring even number of times, but 5 is occurring a single time and we know that a number to be a perfect square its prime factors must be appearing even number of times when written in factorised form. So, we will multiply a 5 to the LCM to get the answer to the above question.
$\text{Required perfect square}=(LCM)\times 5=2\times 2\times 3\times 3\times 5\times 5=900$
 Hence, we can conclude that the answer to the above question is option (d).


Note: If you want you can solve the above question by eliminating the options as well. If you see the options 100 and 400, they are not divisible by 3, as the sum of their digits is not divisible by 3, and as they are not divisible by 3, they cannot be a multiple of 6 as well, so they are eliminated. 256 neither has a 0 or a 5 as its unit digit, so it is not divisible by 5. So, the answer is 900, i.e., option (d).