Euclid’s Division Lemma

Step 1: (23\times22=506) and (23\times23=529). Step 2: The nearest lower multiple of 23 below 511 is 506, so the remainder is 5. Step 3: Do not accept a negative remainder or a remainder greater than the divisor.

Step 1: Number (=) divisor (\times) quotient (+) remainder. Step 2: (45\times16+44=720+44=764). Step 3: The remainder 44 is less than divisor 45, so the form is valid.

Step 1: For the difference, the remainder is (3-9=-6). Step 2: Add 16 to make it valid, giving 10. Step 3: In subtraction, add the divisor when the remainder becomes negative.

Step 1: On division by 6, possible remainders are from 0 to 5. Step 2: Six consecutive integers cover all these remainders once. Step 3: The number with remainder 0 is divisible by 6.

Step 1: The original remainder is 34. Step 2: (71) leaves remainder 1 on division by 35, so total remainder (34+1=35), which becomes 0. Step 3: First reduce the large added number to a small remainder.

Step 1: The remainder of (u) is 5. Step 2: The remainder of (u^2-u) comes from (5^2-5=20), and (20=12+8). Step 3: Substitute the remainder first, then find the final remainder.

Step 1: (7^2=49), which leaves remainder 4 on division by 9. Step 2: For (7^4), use (4^2=16), and 16 leaves remainder 7 on division by 9. Step 3: In higher powers, reduce the remainder after each step.

Step 1: The number is (18q+12). Step 2: Adding 6 gives (18q+18=18(q+1)), so the remainder is 0. Step 3: When the remainder and the added number make the divisor, the new remainder becomes zero.

Step 1: The remainders of (a) and (b) are 9 and 6. Step 2: The remainder of (ab+a) comes from (9\times6+9=63), and (63=14\times4+7). Step 3: In a mixed expression, handle the remainder of each term separately.

Step 1: In standard form, the remainder must be from 0 to 119. Step 2: (120\times10=1200), so (1201=1200+1). Step 3: Along with correct calculation, the valid range of the remainder is necessary.

Step 1: These remainders all represent odd numbers. Step 2: (1^2,3^2,5^2,7^2) all leave remainder 1 when divided by 8. Step 3: Remember that the square of an odd number leaves remainder 1 on division by 8.

Step 1: For eight times the number, the remainder part is (8\times5=40). Step 2: (40=27+13), so the final remainder is 13. Step 3: After multiplication, reduce the result below the divisor.

Step 1: The remainder of (a) is 3. Step 2: The remainder of (a^2+2) comes from (3^2+2=11), and (11=7+4). Step 3: Substitute the remainder in the expression, then find the final remainder.

Step 1: In (3a+b), the remainder part is (3\times6+8=26). Step 2: (26=11\times2+4), so the remainder is 4. Step 3: In multi-term expressions, handle the remainder of each term separately.

Step 1: The original remainder is 41. Step 2: (85) leaves remainder 1 on division by 42, so total remainder (41+1=42), which becomes 0. Step 3: When adding a large number, first find its smaller remainder.

Step 1: Replace (n) by its remainder 3. Step 2: The remainder of (n^2+n+1) comes from (3^2+3+1=13), and 13 leaves remainder 1 when divided by 4. Step 3: In polynomial-like expressions, substituting the remainder makes the solution direct.

Step 1: The remainder of (a) is 23. Step 2: For (6a), compute (6\times23=138), and (138=30\times4+18). Step 3: After multiplication, reduce the answer below the divisor.

Step 1: The remainder must satisfy (0\le r<63). Step 2: 63 is equal to the divisor, so it cannot be a valid remainder. Step 3: In definition-based questions, check the condition (r<b) first.

Step 1: (n^2+2n+1=(n+1)^2). Step 2: Since (n) has remainder 9, (n+1) has remainder 0, so its square is divisible by 10. Step 3: First recognize the structure of the expression to reduce calculation.

Step 1: Add the remainders: (17+19+6=42). Step 2: (42) is exactly divisible by 21, so the remainder should be 0. Step 3: In multi-term questions, add only the remainders and reduce at the end.