Euclid’s Division Lemma

Step 1: (39 \times 2=78) and (39 \times 3=117). Step 2: Since (117) is greater, (q=2) and (r=100-78=22). Step 3: The remainder must be less than (39).

Step 1: Write (a=bq+(b-1)). Step 2: (a+1=bq+b=b(q+1)+0), so the remainder is (0). Step 3: Adding (1) to the greatest remainder brings the remainder back to (0).

Step 1: (a=bq) is exactly divisible by (b). Step 2: (a+1=bq+1), so the remainder is (1). Step 3: Just after an exactly divisible number, the remainder is (1).

Step 1: (28 \times 9=252) and (28 \times 10=280). Step 2: Since (280) is greater, the remainder is (275-252=23). Step 3: If the next multiple is greater, choose the previous multiple.

Step 1: (28 \times 9=252) and (28 \times 10=280). Step 2: Since (280) is greater than (275), the quotient is (9). Step 3: Choose the quotient whose product does not exceed the dividend.

Step 1: Adding (14) to the old remainder (12) gives (26). Step 2: (26=13 \times 2+0), so the new remainder is (0). Step 3: Do not forget to divide the new sum by the same divisor.

Step 1: (a-3=13q-2), but the remainder should not be negative. Step 2: (13q-2=13(q-1)+11), so the remainder is (11). Step 3: When a negative remainder appears, add the divisor to make the correct remainder.

Step 1: In (a=42q+41), the remainder is (41), one less than (42). Step 2: (a+1=42q+42=42(q+1)+0), so the remainder is (0). Step 3: Adding (1) to a (b-1) remainder gives exact division.

Step 1: In (a=42q+40), the remainder is (40). Step 2: (a+5=42q+45=42(q+1)+3), so the remainder is (3). Step 3: If the sum crosses the divisor, subtract the divisor once.

Step 1: On division by (31), the remainder must be from (0) to (30). Step 2: In (31q+30), the remainder is (30), which is less than (31). Step 3: A remainder (31), (37), or negative is not in standard form.

Step 1: Divide the larger number by the smaller number. Step 2: (867=255\times3+102), and (102<255), so the quotient is 3 and the remainder is 102. Step 3: In exams, always check that the remainder is smaller than the divisor.

Step 1: By the division lemma, number (=37\times18+r), where (0\le r<37). Step 2: The greatest remainder is 36, so the number is (666+36=702). Step 3: The greatest remainder is always one less than the divisor.

Step 1: In Euclid’s division lemma, (0\le r<b). Step 2: Here the divisor is 49, so the remainder can only be from 0 to 48. Step 3: If the remainder is equal to or greater than the divisor, reject it immediately.

Step 1: Compare with the standard form (a=bq+r). Step 2: Here the divisor is 17 and the remainder is 16, which is less than 17. Step 3: Always compare the remainder with the divisor to check validity.

Step 1: In Euclidean form, the remainder is non-negative and smaller than the divisor. Step 2: (19\times22=418), so (431=418+13) and (13<19). Step 3: A form with a negative remainder is not the standard Euclidean form.

Step 1: Euclid’s lemma says (0\le r<b). Step 2: Here (b=12), so possible remainders are from 0 to 11. Step 3: Do not forget that 0 is also a possible remainder.

Step 1: When divided by 2, the remainder can only be 0 or 1. Step 2: An odd number is not exactly divisible by 2, so the remainder is 1 and the form is (a=2q+1). Step 3: For even-odd questions, take 2 as the divisor.

Step 1: Match it with (n=5q+r). Step 2: Here (r=3), and (3<5), so it is a valid remainder. Step 3: In such questions, treat the multiple part as the quotient part.

Step 1: The main rule is (a=bq+r). Step 2: The correct condition is (0\le r<b), because the remainder can also be zero. Step 3: Be careful with (0<r), because it excludes exact division.

Step 1: The lemma gives existence as well as uniqueness. Step 2: For fixed (a) and (b), only one valid pair (q,r) satisfies the condition. Step 3: Many algebraic forms may be written, but only the form with a valid remainder is correct.