Euclid’s Division Lemma

Step 1: Remainders start from (0) and go up to one less than the divisor. Step 2: The divisor is (6), so remainders from (0) to (5) are possible. Step 3: The remainder cannot be (6).

Step 1: (r=0) means no remainder is left. Step 2: Therefore (a=bq), so (a) is exactly divisible by (b). Step 3: When the remainder is zero, think of multiples.

Step 1: In (63=7 \times 9+0), the remainder is (0). Step 2: This means (63) is exactly divisible by (7). Step 3: In zero-remainder questions, identify the multiple directly.

Step 1: The remainder must be less than the divisor. Step 2: In (20=6 \times 2+8), remainder (8) is greater than divisor (6). Step 3: It is not enough for the sum to be correct; the remainder range must also be correct.

Step 1: Dividing (23) by (10) gives quotient (2). Step 2: (10 \times 2=20) and (23-20=3). Step 3: Since (3<10), (q=2, r=3) is correct.

Step 1: Apply Euclid’s form (a=bq+r). Step 2: Here (b=10) and (r=7), so the number is (10q+7). Step 3: In such questions, place the divisor and remainder directly in the form.

Step 1: According to the lemma, (a=bq+r). Step 2: Substituting (b=12) and (r=5) gives (a=12q+5). Step 3: The remainder is added and remains less than the divisor.

Step 1: (8 \times 9=72). Step 2: So (72=8 \times 9+0). Step 3: When a number is exactly divisible, the remainder is (0).

Step 1: (9 \times 11=99). Step 2: (100-99=1), so the remainder is (1). Step 3: Using the nearest smaller multiple helps in quick calculation.

Step 1: In (a=bq+r), (a) is the dividend and (b) is the divisor. Step 2: (q) is multiplied by (b), so it is the quotient. Step 3: Identifying symbols reduces mistakes in exams.

Step 1: In (a=bq+r), (r) is added at the end. Step 2: What remains after division is called the remainder. Step 3: Always check (0 \le r < b) for (r).

Step 1: In (a=bq+r), (a) is the number being divided. Step 2: Therefore, (a) is called the dividend. Step 3: Link each symbol with its name to remember it well.

Step 1: In (a=bq+r), (b) is the number by which division is done. Step 2: Therefore, (b) is called the divisor. Step 3: Remember that the divisor cannot be zero.

Step 1: (3 \times 4=12) and (3 \times 5=15). Step 2: (12) is the correct smaller multiple, so the remainder is (14-12=2). Step 3: The remainder (2) is less than (3).

Step 1: (6 \times 9=54). Step 2: (58-54=4), so the remainder is (4). Step 3: Since (4<6), the answer is valid.

Step 1: (6 \times 9=54) and (6 \times 10=60). Step 2: (60) is greater than (58), so the quotient is (9). Step 3: Choose the quotient so that the product does not exceed the dividend.

Step 1: The lemma says that for positive integers, (a=bq+r) can be written. Step 2: The condition (0 \le r < b) is necessary. Step 3: While reading statements, eliminate wrong options using the remainder condition.

Step 1: The Euclidean form is (a=bq+r). Step 2: Taking (b=7) and (r=2), we get (a=7q+2). Step 3: Add the remainder in the correct form instead of subtracting it.

Step 1: Put (b=4) and (r=1) in (a=bq+r). Step 2: This gives (a=4q+1). Step 3: In such forms, (q) can be an integer.

Step 1: (13 \times 7=91). Step 2: So (91=13 \times 7+0). Step 3: When the product is exactly equal to the dividend, the remainder is (0).