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To determine the highest common factor (HCF) of [tex]\(A\)[/tex] and [tex]\(B\)[/tex], we need to follow a series of steps that involve examining their prime factorizations:
Prime Factorizations:
- For [tex]\(A\)[/tex]:
[tex]\[ A = 2^2 \times 3 \times 5 \][/tex]
- For [tex]\(B\)[/tex]:
[tex]\[ B = 2^3 \times 3^2 \times 5 \][/tex]
Finding Common Prime Factors:
1. Prime factor 2:
- In [tex]\(A\)[/tex], the exponent of 2 is 2.
- In [tex]\(B\)[/tex], the exponent of 2 is 3.
- The minimum exponent is 2.
2. Prime factor 3:
- In [tex]\(A\)[/tex], the exponent of 3 is 1.
- In [tex]\(B\)[/tex], the exponent of 3 is 2.
- The minimum exponent is 1.
3. Prime factor 5:
- Both [tex]\(A\)[/tex] and [tex]\(B\)[/tex] have the exponent of 5 as 1.
- The minimum exponent is 1.
Calculating the HCF:
Now that we have the minimum exponents for each common prime factor, we can compute the HCF by multiplying these factors together:
[tex]\[ HCF = 2^{\min(2, 3)} \times 3^{\min(1, 2)} \times 5^{\min(1, 1)} \][/tex]
Simplifying the exponents, we get:
[tex]\[ HCF = 2^2 \times 3^1 \times 5^1 = 4 \times 3 \times 5 \][/tex]
Performing the multiplication:
[tex]\[ 4 \times 3 = 12 \][/tex]
[tex]\[ 12 \times 5 = 60 \][/tex]
Thus, the highest common factor (HCF) of [tex]\(A\)[/tex] and [tex]\(B\)[/tex] is:
[tex]\[ \boxed{60} \][/tex]
Prime Factorizations:
- For [tex]\(A\)[/tex]:
[tex]\[ A = 2^2 \times 3 \times 5 \][/tex]
- For [tex]\(B\)[/tex]:
[tex]\[ B = 2^3 \times 3^2 \times 5 \][/tex]
Finding Common Prime Factors:
1. Prime factor 2:
- In [tex]\(A\)[/tex], the exponent of 2 is 2.
- In [tex]\(B\)[/tex], the exponent of 2 is 3.
- The minimum exponent is 2.
2. Prime factor 3:
- In [tex]\(A\)[/tex], the exponent of 3 is 1.
- In [tex]\(B\)[/tex], the exponent of 3 is 2.
- The minimum exponent is 1.
3. Prime factor 5:
- Both [tex]\(A\)[/tex] and [tex]\(B\)[/tex] have the exponent of 5 as 1.
- The minimum exponent is 1.
Calculating the HCF:
Now that we have the minimum exponents for each common prime factor, we can compute the HCF by multiplying these factors together:
[tex]\[ HCF = 2^{\min(2, 3)} \times 3^{\min(1, 2)} \times 5^{\min(1, 1)} \][/tex]
Simplifying the exponents, we get:
[tex]\[ HCF = 2^2 \times 3^1 \times 5^1 = 4 \times 3 \times 5 \][/tex]
Performing the multiplication:
[tex]\[ 4 \times 3 = 12 \][/tex]
[tex]\[ 12 \times 5 = 60 \][/tex]
Thus, the highest common factor (HCF) of [tex]\(A\)[/tex] and [tex]\(B\)[/tex] is:
[tex]\[ \boxed{60} \][/tex]
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