Euclid’s Division Lemma

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.

Step 1: (27\times11=297) and (27\times12=324). Step 2: The greatest multiple not exceeding 314 is 297, so the remainder is 17. Step 3: Do not accept a negative remainder or a remainder greater than the divisor.

Step 1: The Euclidean form is (a=bq+r). Step 2: If the remainder is 0, then (a=23q+0=23q). Step 3: Remainder 0 means the number is exactly divisible by the divisor.

Step 1: In a valid form, the remainder must be from 0 to 14. Step 2: (15\times6=90), so (98=90+8), and 8 is valid. Step 3: Check both the calculation and the remainder limit.

Step 1: The remainder must satisfy (0\le r<9). Step 2: The possible values are 0 through 8, so there are 9 values. Step 3: The number of possible remainders is equal to the divisor.

Step 1: Check multiples of 64: (64\times15=960). Step 2: (1000-960=40), so the remainder is 40. Step 3: For large numbers, reaching the nearest lower multiple is a fast method.

Step 1: When divided by 4, the remainder can be 0, 1, 2, or 3. Step 2: In (4q+4), the remainder is 4, equal to the divisor, so it is not a standard form. Step 3: A remainder is never equal to the divisor.

Step 1: The number has remainder 5, so the square has the same remainder as (5^2=25) divided by 6. Step 2: (25=6\times4+1), so the remainder is 1. Step 3: In square questions, first square the smaller remainder.

Step 1: Write the number as (7q+6). Step 2: Adding 1 gives (7q+7=7(q+1)), so the remainder is 0. Step 3: If the remainder is one less than the divisor and 1 is added, divisibility becomes exact.

Step 1: The number is (8q+5). Step 2: Adding 11 gives (8q+16=8(q+2)), so the remainder is 0. Step 3: Reduce the added number by the divisor and combine remainders.

Step 1: The remainder of (a) is 9. Step 2: Adding 20 gives total remainder (9+20=29), and (29=13\times2+3). Step 3: When the total remainder exceeds the divisor, reduce it again.

Step 1: Write (x=10q+7). Step 2: (3x=30q+21=10(3q+2)+1), so the remainder is 1. Step 3: In multiplication questions, multiply the remainder and reduce it by the divisor.

Step 1: Let the number be (11q+8). Step 2: Twice the number is (22q+16=11(2q+1)+5), so the remainder is 5. Step 3: The final remainder must always be less than 11.

Step 1: In (a=5q+r), the remainder must satisfy (0\le r<5). Step 2: 5 is equal to the divisor, so it cannot be a remainder. Step 3: Valid remainders start from 0 and end at one less than the divisor.

Step 1: Let (p=6q+4). Step 2: The remainder of (p^2) is the remainder of (4^2=16) divided by 6, and (16=6\times2+4). Step 3: In power-based questions, work with the remainder instead of the whole number.

Step 1: On division by 3, every integer is of the form (3q), (3q+1), or (3q+2). Step 2: Three consecutive integers cover these three remainders, so one is exactly divisible by 3. Step 3: Use the cycle of remainders for consecutive-number problems.