Let's break down the problem step by step, using the given values for [tex]\( X \)[/tex], [tex]\( Y \)[/tex], and [tex]\( Z \)[/tex].
### Given:
- [tex]\( X = 2^8 \)[/tex]
- [tex]\( Y = 2^5 \times 5^3 \)[/tex]
- [tex]\( Z = 2^6 \times 5^2 \times 7 \)[/tex]
### a) The LCM of [tex]\( X, Y \)[/tex], and [tex]\( Z \)[/tex]:
To find the Least Common Multiple (LCM), we need to take the highest power of each prime factor appearing in the factorizations of [tex]\( X, Y \)[/tex], and [tex]\( Z \)[/tex].
- Prime factor 2:
- The highest power is [tex]\( 2^8 \)[/tex] (from [tex]\( X \)[/tex]).
- Prime factor 5:
- The highest power is [tex]\( 5^3 \)[/tex] (from [tex]\( Y \)[/tex]).
- Prime factor 7:
- The highest power is [tex]\( 7^1 \)[/tex] (from [tex]\( Z \)[/tex]).
So, the LCM is the product of the highest powers of the prime factors:
[tex]\[ \text{LCM} = 2^8 \times 5^3 \times 7^1 \][/tex]
### b) The HCF of [tex]\( X, Y \)[/tex], and [tex]\( Z \)[/tex]:
To find the Highest Common Factor (HCF), we need to take the lowest power of each prime factor appearing in the factorizations of [tex]\( X, Y \)[/tex], and [tex]\( Z \)[/tex].
- Prime factor 2:
- The lowest power is [tex]\( 2^5 \)[/tex] (common to [tex]\( Y \)[/tex]).
- Prime factor 5:
- The lowest power is [tex]\( 5^2 \)[/tex] (from [tex]\( Z \)[/tex]).
- Prime factor 7:
- The lowest power is [tex]\( 7^0 \)[/tex] (since [tex]\( X \)[/tex] and [tex]\( Y \)[/tex] do not contain 7, we take [tex]\( 7^0 \)[/tex]).
So, the HCF is the product of the lowest powers of the prime factors:
[tex]\[ \text{HCF} = 2^5 \times 5^2 \times 7^0 \][/tex]
Now, we write our results (numerical values of the exponents were calculated previously):
- The LCM: [tex]\( 2^8 \times 5^3 \times 7^1 = 224000 \)[/tex]
- The HCF: [tex]\( 2^5 \times 5^2 \times 7^0 = 800 \)[/tex]
Therefore:
- The LCM of [tex]\( X, Y \)[/tex], and [tex]\( Z \)[/tex] is [tex]\( 224000 \)[/tex].
- The HCF of [tex]\( X, Y \)[/tex], and [tex]\( Z \)[/tex] is [tex]\( 800 \)[/tex].
