Which of the following is the smallest perfect square divisible by 8 and 15?
Which of the following is the smallest perfect square divisible by 8 and 15?
Here, prime factors 2, 3 and 5 have no pair. Therefore 120 must be multiplied by 2 x 3 x 5 to make it a perfect square. Hence, the smallest square number which is divisible by 8, 15 and 20 is 3600.
What is the square root between 6 and 8?
These numbers are all known as real numbers. All negative numbers will have a complex number as their square root. A complex number is a number multiplied by i. i is the “imaginary” square root of -1….List of Perfect Squares.
| NUMBER | SQUARE | SQUARE ROOT |
|---|---|---|
| 6 | 36 | 2.449 |
| 7 | 49 | 2.646 |
| 8 | 64 | 2.828 |
| 9 | 81 | 3.000 |
What is smallest perfect square number?
Thus, 0 is the smallest perfect square.
How do you write the square root of 6?
The square root of 6 is expressed as √6 in the radical form and as (6)½ or (6)0.5 in the exponent form. The square root of 6 rounded up to 7 decimal places is 2.4494897.
What is a square root between 6 and 7?
The number 40 lies between 36 and 49 hence √40 lies between 6 and 7.
What is the smallest perfect square divisible by 15 21 and 32?
square of 35 is 1225. this is divisible by both 15 and 21.
What is the smallest square number divisible by 12 10 and 15?
The smallest square number divisible by 12, 10 and 15 = 2^2×3^2×5^2 =900. Now let us make up a square number that is common multiple of 12, 10, and 15. Collecting the maximal exponents leads to 2 2 × 3 2 × 5 2 = ( 2 × 3 × 5) 2 = 30 2 = 900.
How do you find the perfect square of 5?
Find the smallest square number that is divisible by each of the numbers 4, 9 and 10. Hence, L.C.M = (2 × 2 × 3 x 3 × 5) = 180. 5 cannot be paired. Therefore, multiply 180 by 5 to get the perfect square.
What is the smallest multiple of 120 that is also a square?
Any number divisible by 6, 8, and 15 is also divisible by the LCM of 6, 8, and 15, which is 120. Thus, we want to find the smallest multiple of 120 that is also a square number, that is, the smallest a such that a 2 = 120 k for some integer k. To do this, first prime factorize 120: 2 3 ⋅ 3 ⋅ 5.