Euclid’s Division Lemma

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).

Step 1: (10 \times 4=40). Step 2: (47-40=7), so the remainder is (7). Step 3: In division by (10), the last digit can quickly show the remainder.

Step 1: (10 \times 4=40) and (10 \times 5=50). Step 2: (50) is greater than (47), so the quotient is (4). Step 3: Decide the quotient using the nearest smaller multiple.

Step 1: The remainder must satisfy (0 \le r < 2). Step 2: So only (r=0) or (r=1) is possible. Step 3: This helps form even and odd numbers.

Step 1: Compare (3q+2) with (a=bq+r). Step 2: Here the divisor is (3) and the remainder is (2). Step 3: Reading the form directly saves time.

Step 1: In (5q+4), (5q) is a multiple of (5). Step 2: The remaining part is (4), so the remainder is (4). Step 3: The remainder is less than the divisor, so the form is valid.

Step 1: The rule for the remainder is (0 \le r < 15). Step 2: Therefore, the greatest possible remainder is (14). Step 3: The greatest remainder is always one less than the divisor.

Step 1: In (a=bq+r), (b) is the divisor. Step 2: In (31=4 \times 7+3), (4) is in the divisor’s place. Step 3: In the product (bq), identify the first factor as the divisor when comparing with the form.

Step 1: In (a=bq+r), (a) is the dividend. Step 2: In the given form, (31) is on the left side, so the dividend is (31). Step 3: The dividend is the number being divided.

Step 1: (7 \times 5=35) and (7 \times 6=42). Step 2: (35) is the correct smaller multiple, so (38-35=3) is the remainder. Step 3: Since (3<7), the form is valid.

Step 1: This lemma shows how to write a number using divisor, quotient, and remainder. Step 2: Its form is (a=bq+r). Step 3: The same idea is also useful later in highest common factor questions.