Match List-I with List-II
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List-I |
List-II |
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[a, b as given in the Euclidean algorithm, quotient (q), Remainder (r)] $a = bq+r$ |
Remainder (r) |
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(A) $a = 112, b = 7$ |
(I) $r=1$ |
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(B) $a = 118, b = 9$ |
(II) $r = 3$ |
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(C) $a = 119, b = 6$ |
(III) $r = 5$ |
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(D) $a = 115, b = 8$ |
(IV) $r = 0$ |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(IV), (B)-(I), (C)-(III), (D)-(II)
|
List-I |
List-II |
|
[a, b as given in the Euclidean algorithm, quotient (q), Remainder (r)] $a = bq+r$ |
Remainder (r) |
|
(A) $a = 112, b = 7$ |
(IV) $r = 0$ |
|
(B) $a = 118, b = 9$ |
(I) $r=1$ |
|
(C) $a = 119, b = 6$ |
(III) $r = 5$ |
|
(D) $a = 115, b = 8$ |
(II) $r = 3$ |
Apply Euclidean division:
(A) $a = 112,\ b = 7$
$112 \div 7 = 16$ remainder $0$ → $r = 0$ ⇒ (A) → (IV)
(B) $a = 118,\ b = 9$
$118 \div 9 = 13$ remainder $1$ → $r = 1$ ⇒ (B) → (I)
(C) $a = 119,\ b = 6$
$119 \div 6 = 19$ remainder $5$ → $r = 5$ ⇒ (C) → (III)
(D) $a = 115,\ b = 8$
$115 \div 8 = 14$ remainder $3$ → $r = 3$ ⇒ (D) → (II)