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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

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Prime Factorization and HCF Calculation

Задание 2 (a)

Задание 2 (b)

Задание 2 (c) (I)

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