120 = 2 * 2 * 2 * 3 * 5 Written as a product of powers is (2^3) * 3 * 5
It is: 23*31*51 = 120
To find the largest odd natural number that is a factor of 120, we first need to factorize 120 into its prime factors, which are 2^3 * 3 * 5. Since we are looking for an odd factor, we can ignore the factor of 2. The largest odd factor of 120 is then 3, as it is the largest odd prime factor present in the prime factorization of 120.
As a product of its prime factors: 2*2*2*3*5 = 120
The prime factors of 120 are: 2, 3, 5
The prime factors of 120 are: 2, 3 and 5. 2 x 2 x 2 x 3 x 5 = 120
It is: 23*31*51 = 120
As a product of its prime factors: 2*2*2*3*5 = 120
The prime factorization of 120 is not 2345 because 2345 is not a factor of 120. The prime factorization of 120 is 2^3 * 3 * 5, which means that 120 can be expressed as the product of primes 2, 3, and 5.
The prime factors of 120 are 2, 3 and 5.
To find the largest odd natural number that is a factor of 120, we first need to factorize 120 into its prime factors, which are 2^3 * 3 * 5. Since we are looking for an odd factor, we can ignore the factor of 2. The largest odd factor of 120 is then 3, as it is the largest odd prime factor present in the prime factorization of 120.
prime factor, 5
As a product of its prime factors: 23*3*5 = 120
As a product of its prime factors: 2*2*2*3*5 = 120 or as 23*3*5 = 120
As a product of its prime factors: 2*2*2*3*5 = 120 or as 23*3*5 = 120
120 60,2 30,2,2 15,2,2,2 5,3,2,2,2
Start by listing out all pairs of numbers greater than 0 and less than or equal to 120 that have a product equal to 120:120 = 1 x 120120 = 2 x 60120 = 3 x 40120 = 4 x 30120 = 5 x 24120 = 6 x 20120 = 8 x 15120 = 10 x 12So, there are 8 factor pairs of 120. Now, show the prime factorization for 120 by expressing it as the product of ALL prime numbers:120 = 2 x 60
Five