Euclid’s Division Lemma

Step 1: On division by 11, possible remainders are from 0 to 10. Step 2: Eleven consecutive integers cover all these remainders once. Step 3: The number with remainder 0 is divisible by 11.

Step 1: The remainders in the two forms are 5 and 7. Step 2: (5^2=25) and (7^2=49), and both leave remainder 1 when divided by 12. Step 3: In form-based questions, work only with the remainder.

Step 1: (a-140=79q+61-140=79q-79). Step 2: This can be written as (79(q-1)+0). Step 3: The number is exactly divisible by 79, so the remainder is 0.

Step 1: For the square, take (27^2=729). Step 2: (729=34\times21+15), so the remainder is 15. Step 3: The final answer must always be less than 34.

Step 1: The remainder of (x) is 13. Step 2: The remainder of (x^2+x) comes from (13^2+13=182). Step 3: Since 182 is exactly divisible by 14, the remainder is 0.

Step 1: (157\times43=6751) and (157\times44=6908). Step 2: The nearest lower multiple below 6789 is 6751, so the remainder is (6789-6751=38). Step 3: Do not accept a negative remainder or a remainder greater than 157.

Step 1: Number (=) divisor (\times) quotient (+) remainder. Step 2: (112\times27+111=3024+111=3135). Step 3: The remainder 111 is less than divisor 112, so the form is valid.

Step 1: For the difference, the remainder is (10-23=-13). Step 2: Add 41 to make it a valid remainder, giving 28. Step 3: In subtraction, add the divisor when the remainder becomes negative.

Step 1: On division by 12, possible remainders are from 0 to 11. Step 2: Twelve consecutive integers cover all these remainders once. Step 3: The number with remainder 0 is divisible by 12.

Step 1: The original remainder is 85. Step 2: 259 leaves remainder 1 when divided by 86. Step 3: The total remainder is (85+1=86), so the final remainder is 0.

Step 1: The remainder of (u) is 13. Step 2: The remainder of (u^2-u) comes from (13^2-13=156). Step 3: (156=20\times7+16), so the remainder is 16.

Step 1: (12^2=144), and 144 leaves remainder 11 when divided by 19. Step 2: For (12^4), check (11^2=121). Step 3: (121=19\times6+7), so the remainder is 7.

Step 1: The original remainder is 37. Step 2: 59 leaves remainder 11 when divided by 48. Step 3: (37+11=48), so the new remainder is 0.

Step 1: The remainders of (a) and (b) are 17 and 22. Step 2: The remainder of (ab+a) comes from (17\times22+17=391). Step 3: (391=24\times16+7), so the final remainder is 7.

Step 1: In standard form, the remainder must be from 0 to 899. Step 2: (900\times10=9000), so (9001=9000+1). Step 3: A negative remainder or a remainder greater than 900 does not make the standard Euclidean form.

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.