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Sagot :
Let's find the Highest Common Factor (H.C.F) of the given sets of numbers using the long division method.
### i. Finding H.C.F of [tex]\(77, 33, 22\)[/tex]:
1. Finding H.C.F of 77 and 33:
- Divide 77 by 33:
[tex]\[ 77 \div 33 = 2 \quad \text{remainder } \ 11 \][/tex]
- Now divide 33 by the remainder 11:
[tex]\[ 33 \div 11 = 3 \quad \text{remainder } \ 0 \][/tex]
- Since the remainder is 0, the H.C.F of 77 and 33 is 11.
2. Now find the H.C.F of 11 and 22:
- Divide 22 by 11:
[tex]\[ 22 \div 11 = 2 \quad \text{remainder } \ 0 \][/tex]
- Since the remainder is 0, the H.C.F of 11 and 22 is 11.
So, the H.C.F of 77, 33, and 22 is 11.
### ii. Finding H.C.F of [tex]\(48, 87, 6\)[/tex]:
1. Finding H.C.F of 48 and 87:
- Divide 87 by 48:
[tex]\[ 87 \div 48 = 1 \quad \text{remainder } \ 39 \][/tex]
- Now divide 48 by the remainder 39:
[tex]\[ 48 \div 39 = 1 \quad \text{remainder } \ 9 \][/tex]
- Now divide 39 by the remainder 9:
[tex]\[ 39 \div 9 = 4 \quad \text{remainder } \ 3 \][/tex]
- Now divide 9 by the remainder 3:
[tex]\[ 9 \div 3 = 3 \quad \text{remainder } \ 0 \][/tex]
- Since the remainder is 0, the H.C.F of 48 and 87 is 3.
2. Now find the H.C.F of 3 and 6:
- Divide 6 by 3:
[tex]\[ 6 \div 3 = 2 \quad \text{remainder } \ 0 \][/tex]
- Since the remainder is 0, the H.C.F of 3 and 6 is 3.
So, the H.C.F of 48, 87, and 6 is 3.
### iii. Finding H.C.F of [tex]\(99, 88\)[/tex]:
1. Finding H.C.F of 99 and 88:
- Divide 99 by 88:
[tex]\[ 99 \div 88 = 1 \quad \text{remainder } \ 11 \][/tex]
- Now divide 88 by the remainder 11:
[tex]\[ 88 \div 11 = 8 \quad \text{remainder } \ 0 \][/tex]
- Since the remainder is 0, the H.C.F of 99 and 88 is 11.
So, the H.C.F of 99 and 88 is 11.
Combining all results:
- The H.C.F of [tex]\(77, 33, 22\)[/tex] is 11
- The H.C.F of [tex]\(48, 87, 6\)[/tex] is 3
- The H.C.F of [tex]\(99, 88\)[/tex] is 11
### i. Finding H.C.F of [tex]\(77, 33, 22\)[/tex]:
1. Finding H.C.F of 77 and 33:
- Divide 77 by 33:
[tex]\[ 77 \div 33 = 2 \quad \text{remainder } \ 11 \][/tex]
- Now divide 33 by the remainder 11:
[tex]\[ 33 \div 11 = 3 \quad \text{remainder } \ 0 \][/tex]
- Since the remainder is 0, the H.C.F of 77 and 33 is 11.
2. Now find the H.C.F of 11 and 22:
- Divide 22 by 11:
[tex]\[ 22 \div 11 = 2 \quad \text{remainder } \ 0 \][/tex]
- Since the remainder is 0, the H.C.F of 11 and 22 is 11.
So, the H.C.F of 77, 33, and 22 is 11.
### ii. Finding H.C.F of [tex]\(48, 87, 6\)[/tex]:
1. Finding H.C.F of 48 and 87:
- Divide 87 by 48:
[tex]\[ 87 \div 48 = 1 \quad \text{remainder } \ 39 \][/tex]
- Now divide 48 by the remainder 39:
[tex]\[ 48 \div 39 = 1 \quad \text{remainder } \ 9 \][/tex]
- Now divide 39 by the remainder 9:
[tex]\[ 39 \div 9 = 4 \quad \text{remainder } \ 3 \][/tex]
- Now divide 9 by the remainder 3:
[tex]\[ 9 \div 3 = 3 \quad \text{remainder } \ 0 \][/tex]
- Since the remainder is 0, the H.C.F of 48 and 87 is 3.
2. Now find the H.C.F of 3 and 6:
- Divide 6 by 3:
[tex]\[ 6 \div 3 = 2 \quad \text{remainder } \ 0 \][/tex]
- Since the remainder is 0, the H.C.F of 3 and 6 is 3.
So, the H.C.F of 48, 87, and 6 is 3.
### iii. Finding H.C.F of [tex]\(99, 88\)[/tex]:
1. Finding H.C.F of 99 and 88:
- Divide 99 by 88:
[tex]\[ 99 \div 88 = 1 \quad \text{remainder } \ 11 \][/tex]
- Now divide 88 by the remainder 11:
[tex]\[ 88 \div 11 = 8 \quad \text{remainder } \ 0 \][/tex]
- Since the remainder is 0, the H.C.F of 99 and 88 is 11.
So, the H.C.F of 99 and 88 is 11.
Combining all results:
- The H.C.F of [tex]\(77, 33, 22\)[/tex] is 11
- The H.C.F of [tex]\(48, 87, 6\)[/tex] is 3
- The H.C.F of [tex]\(99, 88\)[/tex] is 11
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