Euclid’s Division Lemma

Step 1: (7\times12=84). Step 2: The division is exact, so the quotient is (12) and the remainder is (0). Step 3: Strong multiplication tables make such questions quick.

Step 1: Compare with the form (a=bq+r). Step 2: Here the divisor is (15) and the remainder is (14), which is less than (15). Step 3: When the form is already given, the remainder can be read directly.

Step 1: Here the divisor appears to be (7) and the remainder (7). Step 2: The remainder must be less than the divisor, not equal to it. Step 3: If the remainder equals the divisor, it should be carried into the quotient.

Step 1: In (7q+7), the extra (7) can be treated as (7\times1). Step 2: So (7q+7=7(q+1)+0), where the remainder is (0). Step 3: If the remainder equals the divisor, add (1) to the quotient.

Step 1: (5\times13=65). Step 2: (66-65=1), so the remainder is (1). Step 3: For division by (5), the last digit can also help.

Step 1: In (a=bq+r), (a) is the dividend. Step 2: In the given form, (99) is on the left side, so the dividend is (99). Step 3: The dividend is the number being divided.

Step 1: (16\times7=112) and (16\times8=128), which is greater than (120). Step 2: (120-112=8), so the remainder is (8). Step 3: Never choose a multiple greater than the dividend.

Step 1: In (a=4q+r), the divisor is (4). Step 2: The remainder can be (0,1,2,3), but not (4). Step 3: The upper limit of the remainder is one less than the divisor.

Step 1: In (a=9q+0), the remainder is (0). Step 2: A zero remainder means the number is exactly divisible by (9). Step 3: Treat zero remainder as a sign of exact division.

Step 1: The divisor cannot be zero in division. Step 2: In the lemma, (b) is taken as a positive integer. Step 3: Division by zero is not valid, so avoid such options.

Step 1: The condition is (0\le r<10). Step 2: So the smallest possible value of (r) is (0). Step 3: The remainder range always starts from (0).

Step 1: The divisor is (10), so the remainder must be less than (10). Step 2: The greatest integer less than (10) is (9). Step 3: To get the greatest remainder, subtract (1) from the divisor.

Step 1: (11\times4=44). Step 2: (45-44=1), which is less than (11). Step 3: A correct Euclidean form always has the remainder in the valid range.

Step 1: The divisor is (7), so (7q) represents a multiple of it. Step 2: Adding remainder (6) gives (7q+6). Step 3: In such questions, put the divisor in the multiple part.

Step 1: (12q) is exactly divisible by (12). Step 2: The leftover part is (5), and it is less than (12), so it is the remainder. Step 3: In (bq+r), (bq) is the multiple and (r) is the remainder.

Step 1: (9\times11=99). Step 2: (100-99=1), so the quotient is (11) and the remainder is (1). Step 3: Do not accept options with an oversized remainder in Euclid’s lemma.

Step 1: With divisor (3), the remainders can be (0,1,2). Step 2: So there are three different possible remainders. Step 3: For a divisor (b), the number of possible remainders is (b).

Step 1: (4\times6=24) and (4\times7=28). Step 2: Since (28) is greater than (27), the quotient is (6). Step 3: For the quotient, choose a multiple that does not exceed the dividend.

Step 1: We can write (27=4\times6+3). Step 2: So the remainder is (3), and it is less than (4). Step 3: After finding the quotient, subtract to get the remainder.

Step 1: This lemma explains the basic structure of division. Step 2: It helps write the dividend, divisor, quotient, and remainder in the form (a=bq+r). Step 3: In the chapter on real numbers, it becomes a base for later methods.