What is the smallest square number which is divisible by each of the numbers 3, 6, and 15? *
Answers
Answer:
Step-by-step explanation:
1. Using prime factorisation, find the square roots of (a)11025 (b) 4761
11025 = 3 * 3 * 5 * 5 * 7 * 7
11025 = 3² * 5² * 7²
11025 = ( 3* 5 * 7)²
√11025 = 3 * 5 * 7 = 105
4761 = 3 * 3 * 23 * 23
√4761 = 3*23 = 69
2. Using prime factorisation, find the cube roots of (a)512 (b) 2197
512 = 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2
512 = 2³ * 2³ * 2³
512 = 8³
∛512 = 8
2197 = 13 * 13 * 13
∛2197 = 13
3. Is 176 a perfect square? If not, find the smallest number by which it should be multiplied to get a perfect square.
176 = 2 * 2 * 2 * 2 * 11
176 = 2² * 2² * 11
176 is not a perfect square
176 to be multiplied by 11 to get a perfect square
176 * 11 = 2² * 2² * 11²
4. Is 9720 a perfect cube? If not, find the smallest number by which it should be divided to get a perfect cube.
9720 = 2 * 2 * 2 * 3 * 3 * 3 * 3 * 3 * 5
9720 = 2³ * 3³ * 3 * 3 * 5
9720 is not a perfect cube
9720 should be multiplied with 3 * 5 * 5 = 75 to get Perfect cube
9720 * 75 = 2³ * 3³ * 3³ * 5³
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The first number divisible by 8, 15 and 20 will be their LCM i.e 120.
But on Factorising 120 we find that it's factors are not in pairs, whereas pairing is the first rule of finding a square or a square root of any number.
Factors of 120 are 2^3 × 3^1 × 5^1.
Here we see that each prime number needs one more power to become a pair so we increase each of their's exponential powers by one and get 2^4 × 3^2 × 5^2 which is equal to 3600.
And therefore we conclude that the smallest square number divisible by 8, 15 and 20 is 3600, which is the square of 60.