Euclid’s Division Lemma

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.

Step 1: The lemma connects dividend, divisor, quotient, and remainder. Step 2: The correct form is (a=bq+r), where the remainder is at least (0) and less than the divisor. Step 3: Always check the range of the remainder in exams.

Step 1: We can write (35=6\times5+5). Step 2: The quotient is (5) and the remainder is (5), which is less than (6). Step 3: The remainder must never be equal to or greater than the divisor.

Step 1: The range of the remainder is very important in the lemma. Step 2: The remainder may be (0), but it must be less than the divisor (b). Step 3: Read the inequality carefully in such questions.

Step 1: The greatest multiple of (9) not exceeding (47) is (45). Step 2: So (47=9\times5+2), giving (q=5) and (r=2). Step 3: First find the nearest smaller multiple of the divisor.

Step 1: (7\times7=49) and (52-49=3). Step 2: Since (3) is less than (7), (q=7,\ r=3) is correct. Step 3: Reject negative remainders or remainders equal to the divisor.

Step 1: Remainders start from (0) and go up to one less than the divisor. Step 2: The divisor is (8), so possible remainders are (0) to (7). Step 3: Do not include the divisor itself as a possible remainder.

Step 1: In (5k+3), the part (5k) is a multiple of (5). Step 2: A multiple of (5) leaves remainder (0), so the final remainder is (3). Step 3: Learn to identify (r) in the form (bq+r).

Step 1: The lemma gives a systematic way to express division. Step 2: In (a=bq+r), (bq) is a multiple of the divisor and (r) is the remainder. Step 3: In meaning-based questions, understand both the formula and the words.

Step 1: In (a=bq+r), the small added part is the remainder. Step 2: In the given expression, (1) is less than (9), so it is the remainder. Step 3: Do not interchange quotient and remainder.

Step 1: (13\times7=91). Step 2: The number is exactly divisible, so the remainder is (0). Step 3: In exact division, remember to write the remainder as (0).