Find Hcf of 81 and 237 by Euclid’s lemma
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Between 81 and 237; 237 is greater than 81
Division lemma of 237 and 81:
Step 1: 237 = 81 × 2 + 75
Step 2: Since remainder 75 ≠ 0, division lemma is applied to 81 and 75 to get 81 = 75 × 1 + 6
Step 3: Since remainder 6 ≠ 0, division lemma is applied to 75 and 6 to get 75 = 6 × 12 + 3
Step 4: Since remainder 3 ≠ 0, division lemma is applied to 6 and 3 to get 6 = 3 × 2 + 0
The remainder is zero in step 4.
Therefore, the divisor i.e. 3 in this step is the H.C.F. of the given numbers.
The H.C.F. of 237 and 81 is 3
Step 5: From Step 3: 3 = 75 – 6 × 12 -----
From Step 2: 6 = 81 – 75 × 1
Thus, from Step 5, we, get 3 = 75 – (81 – 75 × 1) × 12
⇒ 3 = 75 – (81× 12 – 75 × 12)
Step 6 ⇒ 3 = 75 × 13 – 81× 12
From Step 1, 75 = 237 – 81 × 2
Thus, from Step 6;
3 = (237 – 81 × 2) × 13 – 81× 12
⇒ 3 = (237 × 13 – 81 × 26) – 81× 12
⇒ 3 = 237 × 13 – 81 × 38
⇒ H.C.F. of 237 and 81 = 237 × 13 + 81 × (–38)
237 × 13 + 81 × (–38) is the representation of H.C.F. of 237 and 81 as linear combination of 237 and 81.
Division lemma of 237 and 81:
Step 1: 237 = 81 × 2 + 75
Step 2: Since remainder 75 ≠ 0, division lemma is applied to 81 and 75 to get 81 = 75 × 1 + 6
Step 3: Since remainder 6 ≠ 0, division lemma is applied to 75 and 6 to get 75 = 6 × 12 + 3
Step 4: Since remainder 3 ≠ 0, division lemma is applied to 6 and 3 to get 6 = 3 × 2 + 0
The remainder is zero in step 4.
Therefore, the divisor i.e. 3 in this step is the H.C.F. of the given numbers.
The H.C.F. of 237 and 81 is 3
Step 5: From Step 3: 3 = 75 – 6 × 12 -----
From Step 2: 6 = 81 – 75 × 1
Thus, from Step 5, we, get 3 = 75 – (81 – 75 × 1) × 12
⇒ 3 = 75 – (81× 12 – 75 × 12)
Step 6 ⇒ 3 = 75 × 13 – 81× 12
From Step 1, 75 = 237 – 81 × 2
Thus, from Step 6;
3 = (237 – 81 × 2) × 13 – 81× 12
⇒ 3 = (237 × 13 – 81 × 26) – 81× 12
⇒ 3 = 237 × 13 – 81 × 38
⇒ H.C.F. of 237 and 81 = 237 × 13 + 81 × (–38)
237 × 13 + 81 × (–38) is the representation of H.C.F. of 237 and 81 as linear combination of 237 and 81.
Answered by
1
Between 81 and 237; 237 is greater than 81
Division lemma of 237 and 81:
Step 1: 237 = 81 × 2 + 75
Step 2: Since remainder 75 ≠ 0, division lemma is applied to 81 and 75 to get 81 = 75 × 1 + 6
Step 3: Since remainder 6 ≠ 0, division lemma is applied to 75 and 6 to get 75 = 6 × 12 + 3
Step 4: Since remainder 3 ≠ 0, division lemma is applied to 6 and 3 to get 6 = 3 × 2 + 0
The remainder is zero in step 4.
Therefore, the divisor i.e. 3 in this step is the H.C.F. of the given numbers.
The H.C.F. of 237 and 81 is 3
Step 5: From Step 3: 3 = 75 – 6 × 12 -----
From Step 2: 6 = 81 – 75 × 1
Thus, from Step 5, we, get 3 = 75 – (81 – 75 × 1) × 12
⇒ 3 = 75 – (81× 12 – 75 × 12)
Step 6 ⇒ 3 = 75 × 13 – 81× 12
From Step 1, 75 = 237 – 81 × 2
Thus, from Step 6;
3 = (237 – 81 × 2) × 13 – 81× 12
⇒ 3 = (237 × 13 – 81 × 26) – 81× 12
⇒ 3 = 237 × 13 – 81 × 38
⇒ H.C.F. of 237 and 81 = 237 × 13 + 81 × (–38)
237 × 13 + 81 × (–38) is the representation of H.C.F. of 237 and 81 as linear combination of 237 and 81.
Division lemma of 237 and 81:
Step 1: 237 = 81 × 2 + 75
Step 2: Since remainder 75 ≠ 0, division lemma is applied to 81 and 75 to get 81 = 75 × 1 + 6
Step 3: Since remainder 6 ≠ 0, division lemma is applied to 75 and 6 to get 75 = 6 × 12 + 3
Step 4: Since remainder 3 ≠ 0, division lemma is applied to 6 and 3 to get 6 = 3 × 2 + 0
The remainder is zero in step 4.
Therefore, the divisor i.e. 3 in this step is the H.C.F. of the given numbers.
The H.C.F. of 237 and 81 is 3
Step 5: From Step 3: 3 = 75 – 6 × 12 -----
From Step 2: 6 = 81 – 75 × 1
Thus, from Step 5, we, get 3 = 75 – (81 – 75 × 1) × 12
⇒ 3 = 75 – (81× 12 – 75 × 12)
Step 6 ⇒ 3 = 75 × 13 – 81× 12
From Step 1, 75 = 237 – 81 × 2
Thus, from Step 6;
3 = (237 – 81 × 2) × 13 – 81× 12
⇒ 3 = (237 × 13 – 81 × 26) – 81× 12
⇒ 3 = 237 × 13 – 81 × 38
⇒ H.C.F. of 237 and 81 = 237 × 13 + 81 × (–38)
237 × 13 + 81 × (–38) is the representation of H.C.F. of 237 and 81 as linear combination of 237 and 81.
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