Euclid’s Division Lemma

Step 1: (12 \times 13=156) and (12 \times 14=168). Step 2: (156) is the nearest smaller multiple of (12), so the remainder is (1). Step 3: In the correct answer, the remainder must be less than the divisor and not negative.

Step 1: In Euclid’s division lemma, the remainder may start from (0). Step 2: The remainder is always less than the divisor (21), so (0 \le r < 21) is correct. Step 3: Never take the remainder equal to the divisor.

Step 1: To find the dividend, use (a=bq+r). Step 2: (a=17 \times 11+9=187+9=196). Step 3: Matching words with symbols before calculation reduces mistakes.

Step 1: On division by (13), remainders can be from (0) to (12). Step 2: The total number of these values is (13). Step 3: For a divisor (b), the number of possible remainders is (b).

Step 1: (19 \times 12=228) and (19 \times 13=247). Step 2: Since (247) is greater, the remainder is (246-228=18). Step 3: If the next multiple is greater, use the previous multiple.

Step 1: (24 \times 12=288) and (24 \times 13=312). Step 2: Since (312) is greater, (305=24 \times 12+17) is correct. Step 3: A form with a negative remainder is not the standard Euclidean form.

Step 1: (25) cannot be the remainder because it is greater than (18). Step 2: (25=18+7), so (18q+25=18(q+1)+7). Step 3: Divide a large remainder again by the divisor to bring it into the correct range.

Step 1: (36) is twice (18). Step 2: (18q+36=18(q+2)+0), so the remainder is (0). Step 3: If the added part is a multiple of the divisor, the remainder can become zero.

Step 1: In (a=10q+9), the remainder is (9). Step 2: (a+4=10q+13=10(q+1)+3), so the new remainder is (3). Step 3: If the sum of remainders exceeds the divisor, subtract the divisor.

Step 1: Add (20) to the old remainder (3) to get (23). Step 2: (23=11 \times 2+1), so the new remainder is (1). Step 3: In addition-based questions, divide the new sum by the same divisor again.

Step 1: (35 \times 12=420). Step 2: (437-420=17), so the remainder is (17). Step 3: After getting the answer, check that (17<35).

Step 1: (35 \times 12=420) and (35 \times 13=455). Step 2: Since (455) is greater than (437), the quotient is (12). Step 3: Checking the next multiple helps in choosing the quotient.

Step 1: On division by (8), remainders can be from (0) to (7). Step 2: So in (8q+r), (r=0,1,2,3,4,5,6,7). Step 3: Do not include (8q+8) while writing standard forms.

Step 1: (9 \times 7=63) and (9 \times 8=72). Step 2: (63) is the correct nearest smaller multiple, so the remainder is (64-63=1). Step 3: The remainder must be less than (9).

Step 1: In (a=7q+6), the remainder is (6). Step 2: (a+3=7q+9=7(q+1)+2), so the remainder is (2). Step 3: If the new remainder exceeds the divisor, subtract the divisor to get the answer.

Step 1: Adding (29) to the old remainder (2) gives (31). Step 2: (31=15 \times 2+1), so the new remainder is (1). Step 3: In large addition cases, it is enough to find the remainder of the new sum.

Step 1: In (a=20q+17), the remainder is (17). Step 2: (a-5=20q+12), so the remainder is (12). Step 3: In subtraction-based questions, subtract the given number from the old remainder.

Step 1: (a-5=20q-2), but the remainder cannot be negative. Step 2: (20q-2=20(q-1)+18), so the correct remainder is (18). Step 3: If a negative remainder appears, reduce the quotient by one and make the remainder positive.

Step 1: (23 \times 23=529). Step 2: The number divides exactly, so the remainder is (0). Step 3: Recognizing exact division makes the calculation faster.

Step 1: (23 \times 23=529). Step 2: The product equals the dividend, so the quotient is (23). Step 3: When the remainder is (0), the chosen multiplier is the correct quotient.