Aptitude - Square Root and Cube Root - Discussion

Guys, try to solve such questions by options:


The least perfect square, which is divisible by each of 21, 36 and 66 implies that of all the options correct number should be divisible by 7, 11,2,3.
21=3x7
36=2^2X3^3
66=2x3x11


Of these four options only "1" is divisble by 7.

SO CORRECT ANSWER IS 1.

PS: If there were more than 2 choices divisble by 7, then eliminate one of them by divind them by 11 as well, and so on by the factors of the divisors. It would be your bad luck if you get multiple answers even after eliminating the options..

The question is " least perfect square which is divisible by each of 21, 36 and 66 ".

So take the LCM of the numbers.

And in the LCM we get 2772.

Which can be written as factors as "2 x 2 x 3 x 3 x 7 x 11".

In which 2 and 3 are double in number but 7 and 11 are singles.

For finding the least perfect square we have to multiply the number by 7 and 11 so that we have all the factors double in number i.e.

"2x2x3x3x7x7x11x11".

Whose multiplied number will be"213444".

Which will be a perfect square.

Hi, @Nick! this is the way to take out lcm.

Suppose the number is 27 now,
27will not go with 2 but with 3 so we'll divide it with 3 as follow.
3/27
3/9
3/3
=1
So lcm =3*3*3.

So, in nutshell, we have to take a least no divisible by the given no and then whatever the quotient comes write it in the next line and then again divide that no divisible by least no and then the process continues until you get 1 at last keep solving and in the end write all the multiples together.

Remember it's the least common factor we looking for 21, 36 and 66.

Find the Lcm = 3 * 2 * 7 * 6 * 11 = 3 * 2 * 7 * 3 * 2 * 11 = 2772 (the 3*2 after 7 is a perfect square to replace the 6 that's in the initial equation coz 3*2 is 6).

3 and 2 are the highest common factors while 7 and 11 are the least because they both appear once. So multiply the Lcm (2772) with 7 and 11 and you get your answer;
That is 2772 * 7 * 11 = 213444.

Hope this helped someone understand better.

Example:

3, 9, 21.

Solution: List the prime factors of each.

3:3.
9:3 x 3.
21:3 x 7.

Multiply each factor the greatest number of times it occurs in any of the numbers. 9 has two 3's, and 21 has one 7.

So we multiply 3 two times, and 7 once. This gives us 63, the smallest number that can be divided evenly by 3, 9, and 21.

We check our work by verifying that 63 can be divided evenly by 3, 9, and 21.

Let's solve it by option elimination.

21 divisibility test number should be divisible by 7 and 3,
36 divisibility test number should be divisible by 9 and 4,
66 divisibility test number should be divisible by 11 and 6,
So option A and C are divisible by 11 and B and d are not so discard B and D.
So option C is not a perfect square so the correct answer is Option A.

3|21 36 66
|7 12 22
2|6 11 there fore lcm =3*7*2*6*11=2772
hence these can be written as
3*7*2*2*3*11
here 3 is of two times &2is of two times there fore remaining is 7&11 in order to have all squares 7*11 is taken
ans:213444

We should multiple it by 7 and 11 to make it a square... If you can see in the factors that 2*2 * 3*3 *7 * 11 = 2772 can not make a square until we multiply it by 7 to make 7*7 and by 11 to make 11*11.... Now 2*@ *3*3 * 7*7 *11*11 .... all are sqare ..which means (2*3*7*11)2(sqare) - (462)2(sqare) = 213444 (Ans)

If we divide it by 2 and 3
The figuers will be 2*3*7*11

And to find it's square root is difficult

If there are words like Highest,Biggest then H.C.F. is to be taken
and if there are words like Smallest,least then L.C.F. is to be taken(though this is not an exact interpretation but it has worked in all cases)

| 21 36 66
2| 21 18 33
2| 21 9. 33
3| 7. 3. 11.

==> 2*2*3*3*7*11.
=> 4*9*77,
=> 36*77,
=> 2772.

To make perfect square means, Make all digits square.

Already we have a square for 2 and 3,
Remaining make square i.e., 7^2. 11^2,

So, Totally....2772*11*7=213444.