Find the square root of $63504$.
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Hint: To find the square root of the given number $63504$, we will use prime factorisation method. We will write that number as the multiple of the primes. After that it will be written in the form of a group of two. Then, we will select one prime number from each group and multiply all such prime numbers. The square root of $63504$ will be the product of selected prime numbers.
Complete step by step solution: To solve the given problem, we must know the prime factorisation method. By using the method of prime factorisation, we can express the given number as a product of prime numbers. Therefore, we will write the given number $63504$ as the product of primes. Let us do a prime factorisation of $63504$. Note that here $63504$ is an even number so we can start prime factorisation with prime number $2$.
Therefore, we can write $63504 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 7 \times 7$.Now we will take the same prime numbers together and write them in groups of two as shown below.$63504 = \left( {2 \times 2} \right) \times \left( {2 \times 2} \right) \times \left( {3 \times 3} \right) \times \left( {3 \times 3} \right) \times \left( {7 \times 7} \right)$. Now we will take one number from each group. Therefore, we will get $2,2,3,3$ and $7$ from the first, second, third, fourth and fifth group respectively. Hence, the square root of $63504$ will be the product of these numbers $2,2,3,3$ and $7$.Therefore, $\sqrt {63504} = \sqrt {{2^2} \times {2^2} \times {3^2} \times {3^2} \times {7^2}} $.$ \Rightarrow \sqrt {63504} = 2 \times 2 \times 3 \times 3 \times 7$
$ \Rightarrow \sqrt {63504} = 252$
Therefore, the square root of $63504$ is $252$.
Note: If the number is even then it is divisible by $2$. If the sum of all digits of a number is divisible by $3$ then that number is divisible by $3$. Double the last digit of the number and subtract the doubled number from the remaining number (remaining digits). If the result is divisible by $7$ then that number is divisible by $7$. Note that here we will consider positive differences. In the given problem, $49$ is divisible by $7$ because double of last digit $9$ is $18$ and positive difference of $18$ and remaining number (remaining digit) $4$ is $14$ and the number $14$ is divisible by $7$.
Complete step by step solution: To solve the given problem, we must know the prime factorisation method. By using the method of prime factorisation, we can express the given number as a product of prime numbers. Therefore, we will write the given number $63504$ as the product of primes. Let us do a prime factorisation of $63504$. Note that here $63504$ is an even number so we can start prime factorisation with prime number $2$.
| $2$ | $63504$ |
| $2$ | $31752$ |
| $2$ | $15876$ |
| $2$ | $7938$ |
| $3$ | $3969$ |
| $3$ | $1323$ |
| $3$ | $441$ |
| $3$ | $147$ |
| $7$ | $49$ |
| $7$ | $7$ |
| $1$ |
Therefore, we can write $63504 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 7 \times 7$.Now we will take the same prime numbers together and write them in groups of two as shown below.$63504 = \left( {2 \times 2} \right) \times \left( {2 \times 2} \right) \times \left( {3 \times 3} \right) \times \left( {3 \times 3} \right) \times \left( {7 \times 7} \right)$. Now we will take one number from each group. Therefore, we will get $2,2,3,3$ and $7$ from the first, second, third, fourth and fifth group respectively. Hence, the square root of $63504$ will be the product of these numbers $2,2,3,3$ and $7$.Therefore, $\sqrt {63504} = \sqrt {{2^2} \times {2^2} \times {3^2} \times {3^2} \times {7^2}} $.$ \Rightarrow \sqrt {63504} = 2 \times 2 \times 3 \times 3 \times 7$
$ \Rightarrow \sqrt {63504} = 252$
Therefore, the square root of $63504$ is $252$.
Note: If the number is even then it is divisible by $2$. If the sum of all digits of a number is divisible by $3$ then that number is divisible by $3$. Double the last digit of the number and subtract the doubled number from the remaining number (remaining digits). If the result is divisible by $7$ then that number is divisible by $7$. Note that here we will consider positive differences. In the given problem, $49$ is divisible by $7$ because double of last digit $9$ is $18$ and positive difference of $18$ and remaining number (remaining digit) $4$ is $14$ and the number $14$ is divisible by $7$.