Euclid’s Division Lemma

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.

Step 1: Add 1 to (m=4q+3). Step 2: (m+1=4q+4=4(q+1)), so it is divisible by 4. Step 3: When the remainder 3 gets 1 added, it reaches the next multiple of 4.

Step 1: Number (=) divisor (\times) quotient (+) remainder. Step 2: (18\times24+17=432+17=449). Step 3: Finally check that remainder 17 is less than divisor 18.

Step 1: Number (=21\times31+r), where (0\le r<21). Step 2: For the least value, take (r=0), so the number is (21\times31=651). Step 3: For the least possible number, use remainder zero.

Step 1: The number is (21\times31+r). Step 2: The greatest value of (r) is 20, so the number is (651+20=671). Step 3: In such questions, take the maximum remainder as one less than the divisor.

Step 1: In the lemma, (q) is an integer and (r) is the remainder. Step 2: The key condition is (0\le r<b). Step 3: In definition-based questions, the remainder condition is the most important clue.