Euclid’s Division Lemma

Step 1: (10 \times 12=120). Step 2: (121-120=1), so the remainder is (1). Step 3: In division by (10), the units digit helps find the remainder.

Step 1: (10 \times 12=120) and (10 \times 13=130). Step 2: (130) is greater than (121), so the quotient is (12). Step 3: For the quotient, always take the nearest smaller multiple.

Step 1: The correct remainder must be at least (0) and less than (9). Step 2: In (70=9 \times 7+7), the remainder is (7), which is less than (9). Step 3: Negative or too large remainders are not accepted.

Step 1: (6 \times 10=60) and (6 \times 11=66). Step 2: (60) is the correct smaller multiple and (64-60=4). Step 3: Since (4<6), (64=6 \times 10+4) is the correct form.

Step 1: The number is of the form (11q+10). Step 2: Adding (1) gives (11q+11=11(q+1)). Step 3: Therefore, the new remainder is (0).

Step 1: The remainder range is (0 \le r < b). Step 2: Here (r=21) and (b=21), so the condition (r<b) is not satisfied. Step 3: A remainder is never written equal to the divisor.

Step 1: Compare with the form (a=bq+r). Step 2: (16) is the divisor, (9) is the quotient, and (6) is the remainder. Step 3: Remainder (6) is less than (16), so the form is correct.

Step 1: The original number is (5q+1). Step 2: Adding (4) gives (5q+5=5(q+1)). Step 3: The new number is exactly divisible by (5), so the remainder is (0).

Step 1: (12 \times 6=72) and (12 \times 7=84). Step 2: (72) is the correct smaller multiple and (73-72=1). Step 3: Remainder (1) is less than (12), so the form is correct.

Step 1: The remainder condition is (0 \le r < b). Step 2: When (b=1), we get (0 \le r < 1), so only (r=0) is possible. Step 3: Any integer divided by (1) leaves remainder (0).

Step 1: Dividing (71) by (9), we get (9 \times 7=63). Step 2: (71-63=8), so the quotient is (7) and the remainder is (8). Step 3: Always check that the remainder is smaller than the divisor.

Step 1: In Euclid’s division lemma, the remainder is always greater than or equal to (0) and smaller than the divisor. Step 2: Here the divisor is (12), so (0 \le r < 12). Step 3: Writing (\le 12) is wrong because the remainder cannot equal the divisor.

Step 1: (13 \times 8=104), the closest smaller multiple of (13) to (105). Step 2: (105-104=1), so (q=8) and (r=1). Step 3: In such questions, the remainder must not be negative or greater than the divisor.

Step 1: When a number is divided by (5), the remainder must be smaller than (5). Step 2: The remainder may also be (0), so the possible remainders are (0,1,2,3,4). Step 3: Remember that a remainder is never equal to the divisor.

Step 1: Use the Euclidean form (a=bq+r). Step 2: (a=7 \times 11+4=77+4=81). Step 3: In such questions, multiply the divisor and quotient first, then add the remainder.

Step 1: The remainder must be smaller than (15). Step 2: (15 \times 6=90) and (96-90=6), so the correct form is (96=15 \times 6+6). Step 3: Check not only equality but also the condition on the remainder.

Step 1: On division by (3), the possible remainders are (0,1,2). Step 2: Therefore, the number can be written as (3q+0, 3q+1, 3q+2). Step 3: Build general forms using possible remainders.

Step 1: In Euclid’s division lemma, the remainder satisfies (0 \le r < b). Step 2: Here (b=8), so (r) cannot be equal to (8). Step 3: Remembering the range of the remainder is very important.

Step 1: Divide (43) by (6). Step 2: (6 \times 7=42) and (43-42=1), so (r=1). Step 3: While choosing the quotient, make sure (bq) does not exceed (a).

Step 1: Check multiples of (17). Step 2: (17 \times 7=119) and (17 \times 8=136), which is greater than (126). So the quotient is (7). Step 3: The quotient is the greatest integer for which the product does not exceed the number.