Aptitude - Square Root and Cube Root - Discussion
Discussion Forum : Square Root and Cube Root - General Questions (Q.No. 3)
3.
The least perfect square, which is divisible by each of 21, 36 and 66 is:
Answer: Option
Explanation:
L.C.M. of 21, 36, 66 = 2772.
Now, 2772 = 2 x 2 x 3 x 3 x 7 x 11
To make it a perfect square, it must be multiplied by 7 x 11.
So, required number = 22 x 32 x 72 x 112 = 213444
Discussion:
80 comments Page 4 of 8.
S.manikandan said:
1 decade ago
21----7*3.
36-----3*3*2*2.
66-------3*2*11.
So here, [7*3*3*3*2*2*3*2*11].
As per the rule we multiply 7*11*3*3*2*2 = 2772.
Here both 3 and 2 are doubles. For perfect square all should be double. So multiple with 7*11 we get 213444.
36-----3*3*2*2.
66-------3*2*11.
So here, [7*3*3*3*2*2*3*2*11].
As per the rule we multiply 7*11*3*3*2*2 = 2772.
Here both 3 and 2 are doubles. For perfect square all should be double. So multiple with 7*11 we get 213444.
Achutha reddy said:
1 decade ago
21 = 3*7.
36 = 3*3*2*2.
66 = 11*3*2.
Means our must be divisible by 2, 3, 7, 11.
**All numbers are evens, so 2 is satisfied.
***3 is common factor for 21, 36, 66 so it is also satisfied.
So you have to check for 7 and 11.
i.e 213444.
36 = 3*3*2*2.
66 = 11*3*2.
Means our must be divisible by 2, 3, 7, 11.
**All numbers are evens, so 2 is satisfied.
***3 is common factor for 21, 36, 66 so it is also satisfied.
So you have to check for 7 and 11.
i.e 213444.
Ravi said:
1 decade ago
Simply a no is divided by some other nos that nos factor must be divided that no.
Sandeep said:
1 decade ago
L.C.M: 21, 36, 66 then we have, 2*2*3*3*7*11.
Now square all these term.
Eg: 4*4*9*9*49*121 = 213444.
Now square all these term.
Eg: 4*4*9*9*49*121 = 213444.
Malakarsidda said:
1 decade ago
How to sole this step 2772 = 2 x 2 x 3 x 3 x 7 x 11?
Prashant said:
1 decade ago
For LCM 21, 36 & 66:
3/21 36 66.
2/7 12 22.
3/7. 6. 11.
2/7. 2. 11.
7. 1. 11.
3 x 3 x 2 x 2 x 7 x 11 = 213444.
3/21 36 66.
2/7 12 22.
3/7. 6. 11.
2/7. 2. 11.
7. 1. 11.
3 x 3 x 2 x 2 x 7 x 11 = 213444.
Aditya said:
1 decade ago
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.
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.
Raju said:
1 decade ago
2x2x3x3x7x11 = 2772.
Here all are pair numbers but 7 & 11 are single numbers that's why we multiple by 7 & 11 again.
Here all are pair numbers but 7 & 11 are single numbers that's why we multiple by 7 & 11 again.
Tejaswini said:
1 decade ago
How it is possible?
Santosh said:
1 decade ago
Take that example.
Take LCM of that numbers.
LCM (21, 36, 66) = 2*3*7*2*6*11==> 2*3*7*2*3*11==> 2^2*3^2*11*7.
To make above as perfect square we multiply with 11 and 7.
So (2^2*3^2*7*11)*7*11 = 213444.
Take LCM of that numbers.
LCM (21, 36, 66) = 2*3*7*2*6*11==> 2*3*7*2*3*11==> 2^2*3^2*11*7.
To make above as perfect square we multiply with 11 and 7.
So (2^2*3^2*7*11)*7*11 = 213444.
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