find the smallest square number which is divisible by each of the number 6, 15 and 20
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Answers
Answer:
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.
Step-by-step explanation:
First, let’s factor each of the proposed divisors into its prime factors:
8 = 2 * 2 * 2, 15 = 3 * 5, and 20 = 2 * 2 * 5. The least common multiple (LCM) is then
2 * 2 * 2 * 3 * 5 = 120. Now, any square that is evenly divisible by 8, 15, and 20 has
to be divisible by the LCM and any of its multiples. The multiple we want is the
smallest square, which can be generated by including an even number of each of
the factors in the prime factorization, namely 2 * 2 * 2 (* 2) *3 (* 3) *5 (* 5) or 120
(the LCM) * ( 2 * 3 * 5) = 120 * 30 = 3600.
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