Prime Factorization and HCF Calculation
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Задание 2 (a)
List all the factors of 48.
The factors of 48 are the numbers that divide 48 without leaving a remainder. These are:
1, 2, 3, 4, 6, 8, 12, 16, 24, 48
Answer: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48
Задание 2 (b)
Write 48 as the product of its prime factors.
To find the prime factors of 48, we can use prime factorization:
- 48 = 2 x 24
- 24 = 2 x 12
- 12 = 2 x 6
- 6 = 2 x 3
So, 48 = 2 x 2 x 2 x 2 x 3 = \(2^4 \cdot 3\)
Write 504 as the product of its prime factors.
To find the prime factors of 504, we can use prime factorization:
- 504 = 2 x 252
- 252 = 2 x 126
- 126 = 2 x 63
- 63 = 3 x 21
- 21 = 3 x 7
So, 504 = 2 x 2 x 2 x 3 x 3 x 7 = \(2^3 \cdot 3^2 \cdot 7\)
Answer:
* 48 = \(2^4 \cdot 3\)
* 504 = \(2^3 \cdot 3^2 \cdot 7\)
Задание 2 (c) (I)
Find the highest common factor (HCF) of 48 and 504.
First, list the prime factors of each number:
- 48 = \(2^4 \cdot 3\)
- 504 = \(2^3 \cdot 3^2 \cdot 7\)
To find the HCF, take the lowest power of common prime factors:
- \(2^3\) and \(3^1\)
HCF (48, 504) = \(2^3 \cdot 3 = 8 \cdot 3 = 24\)
Answer: 24
