Euclid’s Division Lemma

Step 1: Euclid’s Division Lemma writes (a=bq+r). Step 2: The remainder is always at least zero and less than the divisor. Step 3: In exams, always check the range of the remainder.

Step 1: Dividing (35) by (6) gives quotient (5). Step 2: (6 \times 5=30) and (35-30=5), so the remainder is (5). Step 3: The remainder is less than (6), so it is valid.

Step 1: The lemma uses dividend (a), divisor (b), quotient (q), and remainder (r). Step 2: The correct relation is (a=bq+r). Step 3: Do not forget the condition (0 \le r < b).

Step 1: When the remainder is (0), the number is exactly divisible. Step 2: Euclid’s form becomes (a=9q+0). Step 3: So the number can be written as (9q).

Step 1: The remainder follows (0 \le r < b). Step 2: Here the divisor is (5), so (r) can be (0) to (4). Step 3: A remainder is never equal to the divisor.

Step 1: The nearest multiple of (7) less than (52) is (49). Step 2: (52-49=3), so the remainder is (3). Step 3: The remainder must be less than (7).

Step 1: In Euclid’s Division Lemma, (a) and (b) are positive integers. Step 2: The divisor (b) cannot be zero, so (b>0). Step 3: In division questions, first check the divisor condition.

Step 1: Dividing (17) by (4) gives quotient (4). Step 2: (4 \times 4=16), leaving remainder (1). Step 3: Since (1<4), this is the correct Euclidean form.

Step 1: The divisor is (3), so possible remainders are (0,1,2). Step 2: There are (3) such remainders. Step 3: For divisor (b), there are (b) possible remainders.

Step 1: When divided by (2), the remainder can only be (0) or (1). Step 2: Even numbers leave (0), while odd numbers leave (1). Step 3: Write an odd number as (2q+1).

Step 1: An even number is exactly divisible by (2). Step 2: So the remainder is (0), giving (a=2q+0). Step 3: This simplifies to (2q).

Step 1: On division by (2), the remainder can be (0) or (1). Step 2: An odd number leaves remainder (1). Step 3: Therefore, an odd number is written as (2q+1).

Step 1: Compare with (a=bq+r). Step 2: In (81=10 \times 8+1), (q=8). Step 3: The number multiplied with the divisor is the quotient.

Step 1: In (a=bq+r), the number added at the end is the remainder. Step 2: Here the added number is (1). Step 3: Since (1<10), the remainder is valid.

Step 1: The remainder must satisfy (0 \le r < 4). Step 2: Possible remainders are (0,1,2,3). Step 3: The greatest remainder is (3), not (4).

Step 1: The divisor is (11), so (r<11). Step 2: Possible remainders are (0) to (10). Step 3: The greatest remainder is always one less than the divisor.

Step 1: The remainder is always less than the divisor. Step 2: Here the remainder is (8) and the divisor is also (8), so it is not valid. Step 3: In such questions, apply (r<b) immediately.

Step 1: The nearest multiple of (5) less than (29) is (25). Step 2: (29-25=4), so (q=5) and (r=4). Step 3: Since (4<5), the form is valid.

Step 1: (9 \times 4=36) and (9 \times 5=45). Step 2: (45) is greater than (44), so the quotient is (4). Step 3: While choosing the quotient, do not take a multiple greater than the dividend.

Step 1: (9 \times 4=36). Step 2: (44-36=8), so the remainder is (8). Step 3: Since (8<9), the remainder is in the correct range.