Question

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?

Answer 1

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.