Euclid’s Division Lemma

Step 1: The possible remainders are 1 and 5. Step 2: (1^2=1) and (5^2=25=6\times4+1), so the remainder is 1 in both cases. Step 3: Squares of many numbers coprime to 6 show remainder 1, but checking is always better.

Step 1: For seven times the number, the remainder part is (7\times4=28). Step 2: (28=22+6), so the final remainder is 6. Step 3: If the result after multiplication is greater than the divisor, reduce it.

Step 1: The remainder of (a) is 2. Step 2: The remainder of (a^2+1) comes from (2^2+1=5), which is exactly divisible by 5. Step 3: After substituting the remainder in the expression, check it again by the divisor.

Step 1: In (2a+b), the remainder part is (2\times4+7=15). Step 2: (15=9+6), so the remainder is 6. Step 3: In such questions, handle the remainder of each term separately.

Step 1: The original remainder is 29. Step 2: 61 leaves remainder 1 when divided by 30, so total remainder (29+1=30), which becomes 0. Step 3: It is useful to reduce a large added number to a smaller remainder first.

Step 1: Replace (n) by its remainder 2. Step 2: The remainder of (n^2+n+1) comes from (2^2+2+1=7), and 7 leaves remainder 1 when divided by 3. Step 3: In polynomial-like expressions, substituting the remainder makes the solution simple.

Step 1: The remainder of (a) is 17. Step 2: For (5a), compute (5\times17=85), and (85=24\times3+13). Step 3: After multiplication, the remainder must be reduced below the divisor.

Step 1: The remainder must satisfy (0\le r<41). Step 2: 41 is equal to the divisor, so it cannot be a remainder. Step 3: In definition-based questions, (r<b) is the most important rule.

Step 1: The expression (n^2+2n+1) is ((n+1)^2). Step 2: Since (n) has remainder 7, (n+1) has remainder 0, so its square is divisible by 8. Step 3: Recognizing the structure of the expression reduces calculation.

Step 1: Add the remainders: (13+16+7=36). Step 2: 36 is exactly divisible by 18, so the remainder is 0. Step 3: In multi-term questions, add only the remainders and then reduce at the end.

Step 1: Find the nearest lower multiple of 112 below 1365. Step 2: (112\times12=1344), so the remainder is (1365-1344=21). Step 3: In exams, always check that the final remainder is smaller than the divisor.

Step 1: The number has the form (39\times22+r), where (0\le r<39). Step 2: The greatest remainder is 38, so the number is (858+38=896). Step 3: For the greatest number, take the remainder one less than the divisor.

Step 1: The number is (57\times18+r). Step 2: For the least value, (r=0), so the number is (57\times18=1026). Step 3: For a minimum value, start with remainder zero.

Step 1: A valid remainder must be from 0 to 72. Step 2: (73\times12=876), so (947=876+71), and 71 is valid. Step 3: A form with a negative remainder may look close, but it is not Euclidean form.

Step 1: The remainder of (x) is 25. Step 2: Adding 28 gives total remainder (25+28=53), and (53=26\times2+1). Step 3: After addition, reduce the remainder again by the divisor.

Step 1: Let (n=19q+8). Step 2: The remainder of (5n+7) comes from (5\times8+7=47), and (47=19\times2+9). Step 3: In a linear expression, using the remainder keeps the calculation short.

Step 1: The square remainder comes from dividing (9^2=81) by 11. Step 2: (81=11\times7+4), so the remainder is 4. Step 3: In square questions, square the remainder instead of the whole number.

Step 1: On division by 9, possible remainders are from 0 to 8. Step 2: Therefore, all forms are from (9q) to (9q+8). Step 3: Include remainder 0 and do not include remainder 9.

Step 1: The remainder of (a) is 30, one less than 31. Step 2: Adding 1 gives (31q+31=31(q+1)), so the remainder is 0. Step 3: Adding 1 to a remainder one less than the divisor gives the next exact multiple.

Step 1: Write (m=17q+4). Step 2: (m-23=17q-19=17(q-2)+15), so the remainder is 15. Step 3: If subtraction gives a negative remainder, add the divisor as needed.