Answer
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Hint: In this problem, first we will find the prime factorization of a given number $ 600 $ . To find the total number of divisors, we will multiply the power of each prime factor by adding $ 1 $ . Then, we will find the number of odd divisors by considering only odd prime factors. To find the number of even divisors we will subtract the number of odd divisors from the total number of divisors.
Complete step-by-step answer:
Here we need to find the number of odd and even divisors of $ 600 $ . First we will find the prime factorization of this number. $ 600 $ is an even number so we can start the prime factorization with number $ 2 $ .
Therefore,
$ \Rightarrow 600 = 2 \times 2 \times 2 \times 3 \times 5 \times 5 = {2^3} \times {3^1} \times {5^2} $ .
Here we can see that the prime factorization of $ 600 $ is of the form $ {p^a} \times {q^b} \times {r^c} $ where $ p = 2 $ , $ q = 3 $ , $ r = 5 $ are prime numbers and $ a = 3 $ , $ b = 1 $ , $ c = 2 $ are powers of prime numbers $ p $ , $ q $ , $ r $ respectively.
Now we are going to find the total number of divisors. For this, we will multiply the power of each prime factor by adding $ 1 $ . Therefore, total number of divisors $ = \left( {a + 1} \right)\left( {b + 1} \right)\left( {c + 1} \right) = \left( {3 + 1} \right)\left( {1 + 1} \right)\left( {2 + 1} \right) = 4 \times 2 \times 3 = 24 $ .
Now we are going to find the number of odd divisors. For this, we will multiply the power of only odd prime factors by adding $ 1 $ . Therefore, number of odd divisors $ = \left( {b + 1} \right)\left( {c + 1} \right) = \left( {1 + 1} \right)\left( {2 + 1} \right) = 2 \times 3 = 6 $ . Now we are going to find the number of even divisors. For this, we will subtract the number of odd divisors from the total number of divisors. Therefore, number of even divisors $ = 24 - 6 = 18 $ .
So, the correct answer is “EVEN DIVISOR= 18, ODD DIVISOR= 6”.
Note: Every integer greater than $ 1 $ either is a prime number itself or can be written as a product of prime numbers. This theorem is called the fundamental theorem of arithmetic or unique factorization theorem. The total number of divisors is denoted by $ d\left( n \right) $ . If the sum of digits of a number is divisible by $ 3 $ then that number is divisible by $ 3 $ . If the unit digit of a number is $ 0 $ or $ 5 $ then that number is divisible by $ 5 $ .
Complete step-by-step answer:
Here we need to find the number of odd and even divisors of $ 600 $ . First we will find the prime factorization of this number. $ 600 $ is an even number so we can start the prime factorization with number $ 2 $ .
$ 2 $ | $ 600 $ |
$ 2 $ | $ 300 $ |
$ 2 $ | $ 150 $ |
$ 3 $ | $ 75 $ |
$ 5 $ | $ 25 $ |
$ 5 $ | $ 5 $ |
$ 1 $ |
Therefore,
$ \Rightarrow 600 = 2 \times 2 \times 2 \times 3 \times 5 \times 5 = {2^3} \times {3^1} \times {5^2} $ .
Here we can see that the prime factorization of $ 600 $ is of the form $ {p^a} \times {q^b} \times {r^c} $ where $ p = 2 $ , $ q = 3 $ , $ r = 5 $ are prime numbers and $ a = 3 $ , $ b = 1 $ , $ c = 2 $ are powers of prime numbers $ p $ , $ q $ , $ r $ respectively.
Now we are going to find the total number of divisors. For this, we will multiply the power of each prime factor by adding $ 1 $ . Therefore, total number of divisors $ = \left( {a + 1} \right)\left( {b + 1} \right)\left( {c + 1} \right) = \left( {3 + 1} \right)\left( {1 + 1} \right)\left( {2 + 1} \right) = 4 \times 2 \times 3 = 24 $ .
Now we are going to find the number of odd divisors. For this, we will multiply the power of only odd prime factors by adding $ 1 $ . Therefore, number of odd divisors $ = \left( {b + 1} \right)\left( {c + 1} \right) = \left( {1 + 1} \right)\left( {2 + 1} \right) = 2 \times 3 = 6 $ . Now we are going to find the number of even divisors. For this, we will subtract the number of odd divisors from the total number of divisors. Therefore, number of even divisors $ = 24 - 6 = 18 $ .
So, the correct answer is “EVEN DIVISOR= 18, ODD DIVISOR= 6”.
Note: Every integer greater than $ 1 $ either is a prime number itself or can be written as a product of prime numbers. This theorem is called the fundamental theorem of arithmetic or unique factorization theorem. The total number of divisors is denoted by $ d\left( n \right) $ . If the sum of digits of a number is divisible by $ 3 $ then that number is divisible by $ 3 $ . If the unit digit of a number is $ 0 $ or $ 5 $ then that number is divisible by $ 5 $ .
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