Euclid’s Division Lemma

Step 1: Check the squares of the given remainders. Step 2: (1^2) and (13^2) give 1, (3^2) and (11^2) give 9, and (5^2) and (9^2) give 11 as remainders. Step 3: Write possible remainders without repeating them.

Step 1: For fifteen times the number, the remainder part is (15\times14=210). Step 2: (210=57\times3+39), so the final remainder is 39. Step 3: After multiplication, divide the result again by the divisor.

Step 1: The remainder of (a) is 11. Step 2: The remainder of (a^2+9) comes from (11^2+9=130). Step 3: (130=18\times7+4), so the final remainder is 4.

Step 1: In (6a+5b), the remainder part is (6\times14+5\times19=179). Step 2: (179=23\times7+18), so the remainder is 18. Step 3: In multi-term expressions, handle each term’s remainder separately.

Step 1: The original remainder is 93. Step 2: 283 leaves remainder 1 when divided by 94. Step 3: (93+1=94), so the final remainder is 0.

Step 1: Replace (n) by its remainder 6. Step 2: The remainder of (n^2+n+1) comes from (6^2+6+1=43). Step 3: Dividing 43 by 7 gives remainder 1.

Step 1: The remainder of (a) is 41. Step 2: For (12a), check (12\times41=492). Step 3: (492=54\times9+6), so the remainder is 6.

Step 1: The remainder must satisfy (0\le r<121). Step 2: 121 is equal to the divisor, so it cannot be a valid remainder. Step 3: In definition questions, check the remainder range first.

Step 1: (n^2+2n+1=(n+1)^2). Step 2: Since the remainder of (n) is 25, the remainder of (n+1) is 0. Step 3: When (n+1) is divisible by 26, its square is also divisible by 26.

Step 1: Add the remainders: (42+45+7=94). Step 2: 94 is exactly divisible by 47. Step 3: Therefore, the final remainder is 0; adding only the remainders is a quick method for multi-term expressions.