Euclid’s Division Lemma

Step 1: (9 \times 8=72) and (9 \times 9=81). Step 2: (76-72=4), so the quotient is (8) and the remainder is (4). Step 3: In exams, finally check that the remainder is less than the divisor.

Step 1: According to Euclid’s Division Lemma, (a=bq+r). Step 2: Here the divisor is (6) and the remainder is (4), so the form is (6q+4). Step 3: In such questions, identify the divisor and remainder directly.

Step 1: (12 \times 7=84) and (12 \times 8=96). Step 2: The nearest smaller multiple of (12) is (84), so (95-84=11). Step 3: Since (11<12), remainder (11) is valid.

Step 1: (5 \times 8=40). Step 2: (43-40=3), so (43=5 \times 8+3) is correct. Step 3: It is not enough for the sum to match; the remainder must be less than (5).

Step 1: The rule for the remainder is (0 \le r < b). Step 2: Here (b=13), so the greatest remainder is (12). Step 3: The greatest possible remainder is always one less than the divisor.

Step 1: Remainders start from zero and go up to one less than the divisor. Step 2: On division by (4), (0,1,2,3) are possible. Step 3: Remainder (4) is not possible because it equals the divisor.

Step 1: In (a=bq+r), (b) is the divisor. Step 2: In (67=8 \times 8+3), (8) is in the divisor’s place. Step 3: Identifying symbols helps solve short questions quickly.

Step 1: In Euclidean form (a=bq+r), the number added at the end is (r). Step 2: In the given form, (3) is added at the end. Step 3: Since (3<8), the remainder is in the correct range.

Step 1: Compare (9q+6) with (a=bq+r). Step 2: Here the divisor is (9) and the remainder is (6). Step 3: Identifying the remainder from the form saves time in exams.

Step 1: The remainder must always be less than the divisor. Step 2: In (51=7 \times 6+9), remainder (9) is greater than divisor (7). Step 3: In such options, check both the sum and the remainder range.

Step 1: On division by (2), the remainder can be (0) or (1). Step 2: So the number becomes (2q+0) or (2q+1). Step 3: These forms identify even and odd numbers.

On dividing 112 by 15, \(15 \times 7=105\) and \(112-105=7\). Thus, \(112=15 \times 7+7\), so the remainder is 7. A remainder of 15 is not possible because a remainder must always be less than the divisor. Exam tip: In \(a=bq+r\), always check that \(0\le r<b\).

Step 1: (15 \times 7=105) and (15 \times 8=120). Step 2: (120) is greater than (112), so the quotient is (7). Step 3: While choosing the quotient, do not take a multiple greater than the dividend.

Step 1: (11q) is a multiple of (11). Step 2: Dividing a multiple by the same number leaves remainder (0). Step 3: In multiple forms, identify zero remainder quickly.

Step 1: The remainder range is (0 \le r < 18). Step 2: Therefore, the greatest possible remainder is (17). Step 3: In this type of question, the answer is one less than the divisor.

Step 1: (10 \times 8=80) and (10 \times 9=90). Step 2: (90) is greater than (89), so we take (80). Step 3: (89-80=9), so the correct form is (89=10 \times 8+9).

Step 1: The remainder must satisfy (0 \le r < b). Step 2: When (b=20), (r) must be less than (20). Step 3: (20) equals the divisor, so it cannot be a remainder.

Step 1: (7q+0) simplifies to (7q). Step 2: It is a multiple of (7), so it is exactly divisible by (7). Step 3: Zero remainder means exact division.

Step 1: The original number is of the form (5q+3). Step 2: Adding (2) gives (5q+5=5(q+1)). Step 3: This is exactly divisible by (5), so the remainder is (0).

Step 1: The number is of the form (8q+5). Step 2: Adding (3) gives (8q+8=8(q+1)). Step 3: Now the number is exactly divisible by (8), so the remainder is (0).