Euclid’s Division Lemma

Step 1: Compare with the form (a=bq+r). Step 2: The number multiplied with (9) is (8), so the quotient is (8). Step 3: The quotient is written as the multiplier of the divisor.

Step 1: The remainder cannot be (6) because it equals the divisor. Step 2: (6q+6=6(q+1)+0), so the correct remainder is (0). Step 3: When the remainder seems equal to the divisor, increase the quotient by one.

Step 1: On division by (7), remainders can be from (0) to (6). Step 2: There are (7) possible remainders in total. Step 3: For a divisor (b), the number of possible remainders is (b).

Step 1: (n=2q) means (n) is exactly divisible by (2). Step 2: A number divisible by (2) is an even number. Step 3: Remember the form (2q) for identifying even numbers.

Step 1: In (2q+1), division by (2) leaves remainder (1). Step 2: Such a number is an odd number. Step 3: Every odd number can be written in the form (2q+1).

Step 1: (16 \times 9=144). Step 2: (150-144=6), so (150=16 \times 9+6). Step 3: The remainder (6) is less than (16), so the form is correct.

Step 1: (15) cannot be the remainder because it is greater than (10). Step 2: (15=10+5), so (10q+15=10(q+1)+5). Step 3: The correct remainder is always less than the divisor.

Step 1: When (r=0), the form becomes (a=bq). Step 2: This means (a) is exactly divisible by (b). Step 3: Zero remainder is a sign of exact divisibility.

Step 1: (17 \times 12=204). Step 2: (17 \times 13=221), which is greater than (208), so the quotient is (12). Step 3: While choosing the quotient, also check the next multiple.

Step 1: (17 \times 12=204). Step 2: (208-204=4), so the remainder is (4). Step 3: Check that the remainder (4<17).

Step 1: The remainder must always be less than (9). Step 2: The greatest integer less than (9) is (8). Step 3: The greatest remainder is (b-1).

Step 1: Dividing a number by itself gives quotient (1). Step 2: Nothing remains, so (35=35 \times 1+0). Step 3: When the dividend and divisor are equal, the remainder is (0).

Step 1: Write (n=5q+3). Step 2: (n+1=5q+4), so the new remainder is (4). Step 3: For a small increase, first add it to the old remainder.

Step 1: Write (n=5q+4). Step 2: (n+1=5q+5=5(q+1)+0). Step 3: When the old remainder is one less than the divisor and (1) is added, the new remainder becomes (0).

Step 1: (6 \times 10=60). Step 2: (64-60=4), so (64=6 \times 10+4) is correct. Step 3: The remainder (4) is less than the divisor (6).

Step 1: Compare with (a=bq+r). Step 2: In (a=4q+2), (r=2). Step 3: Since (2<4), this remainder is valid.

Step 1: Euclid’s division lemma is applied to two positive integers. Step 2: (b) is the divisor and it cannot be zero. Step 3: For the dividing number, positivity and non-zero value are necessary.

Step 1: (14 \times 8=112). Step 2: (119-112=7), so the remainder is (7). Step 3: Do not forget to check that the final remainder is less than the divisor.

Step 1: (9) cannot be the remainder because it is greater than (7). Step 2: (9=7+2), so (7q+9=7(q+1)+2). Step 3: A large remainder must be converted into the correct range.

Step 1: The main condition on the remainder is (0 \le r < b). Step 2: This means the remainder is less than the divisor. Step 3: In theory-based questions, remember this condition directly.

Step 1: In the form (a=bq+r), (b) is the divisor. Step 2: Here (13) is multiplied by (6), so (13) is the divisor. Step 3: While identifying terms, look at the first number in the product.

Step 1: The Euclidean form is (a=bq+r). Step 2: Here (b=10) and (r=6), so the number is (10q+6). Step 3: Put the divisor and the remainder in the correct places.

Step 1: The dividend (49) is smaller than the divisor (50). Step 2: The quotient is (0) and the remainder remains (49). Step 3: When a smaller number is divided by a larger number, the remainder can be the smaller number itself.

Step 1: The remainder is less than the divisor. Step 2: The greatest integer less than (18) is (17). Step 3: If (b) is given, the greatest remainder is (b-1).

Step 1: Write (n=8q+3). Step 2: (n+4=8q+7), so the remainder is (7). Step 3: Add the added number to the old remainder and check whether the sum is less than the divisor.