as a product of its prime factors,
120=2x2x2x3x5
How could you use this information to find out all of the factors of 120, making sure that you do not miss any factor pairs?
To find all the factors of 120, we need to consider every possible combination of the prime factors (2, 2, 2, 3, 5) and their exponents.
1. Start by listing all the possible powers of 2: 2^0, 2^1, 2^2, 2^3.
2. For each power of 2, consider all possible powers of 3: 3^0, 3^1.
3. Finally, for each combination of exponents for 2 and 3, consider all possible powers of 5: 5^0, 5^1.
The factors of 120 are obtained by multiplying the different combinations of prime factors and their exponents. Here are all the factor pairs of 120:
2^0 * 3^0 * 5^0 = 1
2^1 * 3^0 * 5^0 = 2
2^2 * 3^0 * 5^0 = 4
2^3 * 3^0 * 5^0 = 8
2^0 * 3^1 * 5^0 = 3
2^1 * 3^1 * 5^0 = 6
2^2 * 3^1 * 5^0 = 12
2^3 * 3^1 * 5^0 = 24
2^0 * 3^0 * 5^1 = 5
2^1 * 3^0 * 5^1 = 10
2^2 * 3^0 * 5^1 = 20
2^3 * 3^0 * 5^1 = 40
2^0 * 3^1 * 5^1 = 15
2^1 * 3^1 * 5^1 = 30
2^2 * 3^1 * 5^1 = 60
2^3 * 3^1 * 5^1 = 120
Hence, these are all the factor pairs of 120, ensuring that none are missed.
To find all the factors of 120, we can use the prime factorization method. The prime factorization of 120 is 2^3 x 3 x 5, which means the prime factors of 120 are 2, 2, 2, 3, and 5.
To determine all the factors of 120, we can derive combinations of these prime factors. We can do this by listing down all possible ways to select and multiply the prime factors, making sure not to miss any combinations.
Starting with the first prime factor, 2, we can have the following combinations with different powers of 2:
- 2^0 = 1 (choosing 2 zero times)
- 2^1 = 2 (choosing 2 once)
- 2^2 = 4 (choosing 2 twice)
- 2^3 = 8 (choosing 2 three times)
Next, we move on to the next prime factor, 3:
- 3^0 = 1 (choosing 3 zero times)
- 3^1 = 3 (choosing 3 once)
Finally, we consider the last prime factor, 5:
- 5^0 = 1 (choosing 5 zero times)
Now we can multiply the combinations of these prime factors to find all the factors of 120:
1 x 1 x 1 = 1
2 x 1 x 1 = 2
4 x 1 x 1 = 4
8 x 1 x 1 = 8
1 x 3 x 1 = 3
2 x 3 x 1 = 6
4 x 3 x 1 = 12
8 x 3 x 1 = 24
1 x 1 x 5 = 5
2 x 1 x 5 = 10
4 x 1 x 5 = 20
8 x 1 x 5 = 40
Therefore, the factors of 120 are: 1, 2, 3, 4, 5, 6, 8, 10, 12, 20, 24, 40, 60, and 120.
To find all the factors of 120, we can start by listing the powers of each prime factor: 2^3, 3^1, and 5^1.
Next, we can generate all possible combinations of these prime factors, considering different powers for each prime factor from 0 to their respective powers.
For example, for the prime factor 2, we can have combinations with powers 0, 1, 2, and 3. Similarly, for the prime factors 3 and 5, we can have combinations with powers 0 and 1.
Now, let's calculate all the combinations:
1. 2^0 * 3^0 * 5^0 = 1 (corresponding to the factor 1)
2. 2^1 * 3^0 * 5^0 = 2
3. 2^2 * 3^0 * 5^0 = 4
4. 2^3 * 3^0 * 5^0 = 8
5. 2^0 * 3^1 * 5^0 = 3
6. 2^1 * 3^1 * 5^0 = 6
7. 2^2 * 3^1 * 5^0 = 12
8. 2^3 * 3^1 * 5^0 = 24
9. 2^0 * 3^0 * 5^1 = 5
10. 2^1 * 3^0 * 5^1 = 10
11. 2^2 * 3^0 * 5^1 = 20
12. 2^3 * 3^0 * 5^1 = 40
These combinations give us all possible factors of 120. Therefore, the factors of 120 are: 1, 2, 3, 4, 5, 6, 8, 10, 12, 20, 24, 40, 60, and 120.