Euclid’s Division Lemma

Step 1: The possible remainders on division by 7 are 0, 1, 2, 3, 4, 5, and 6. Step 2: Seven consecutive integers cover all these remainders once. Step 3: The number with remainder 0 is divisible by 7.

Step 1: The remainders in the two forms are 3 and 5. Step 2: (3^2=9) and (5^2=25), and both leave remainder 1 on division by 8. Step 3: Checking squares of odd remainders modulo 8 is a quick method.

Step 1: (a-103=59q+44-103=59q-59). Step 2: This can be written as (59(q-1)+0). Step 3: Therefore, the number is exactly divisible by 59 and the remainder is 0.

Step 1: For the square, take (17^2=289). Step 2: (289=21\times13+16), so the remainder is 16. Step 3: The final answer must be smaller than 21, so 289 cannot be the answer.

Step 1: The remainder of (x) is 9. Step 2: The remainder of (x^2+x) comes from (9^2+9=90). Step 3: Since 90 is exactly divisible by 10, the remainder is 0.

Step 1: (89\times22=1958) and (89\times23=2047). Step 2: The nearest lower multiple below 2026 is 1958, so the remainder is (2026-1958=68). Step 3: Do not accept a negative remainder or a remainder greater than 89.

Step 1: Number (=) divisor (\times) quotient (+) remainder. Step 2: (76\times24+75=1824+75=1899). Step 3: The remainder 75 is less than divisor 76, so this form is valid.

Step 1: For the difference, the remainder is (5-14=-9). Step 2: Add 27 to make it a valid remainder, giving 18. Step 3: In subtraction, add the divisor when the remainder becomes negative.

Step 1: On division by 8, possible remainders are from 0 to 7. Step 2: Eight consecutive integers cover all these remainders once. Step 3: The number with remainder 0 is divisible by 8.

Step 1: The original remainder is 43. Step 2: 133 leaves remainder 1 when divided by 44. Step 3: The total remainder is (43+1=44), so the final remainder is 0.

Step 1: The remainder of (u) is 8. Step 2: The remainder of (u^2-u) comes from (8^2-8=56). Step 3: (56=15\times3+11), so the remainder is 11.

Step 1: (7^2=49), and 49 leaves remainder 10 when divided by 13. Step 2: For (7^4), check (10^2=100). Step 3: (100=13\times7+9), so the remainder is 9.

Step 1: The original remainder is 17. Step 2: 31 leaves remainder 7 when divided by 24. Step 3: (17+7=24), so the new remainder is 0.

Step 1: The remainders of (a) and (b) are 11 and 14. Step 2: The remainder of (ab+a) comes from (11\times14+11=165). Step 3: (165=16\times10+5), so the final remainder is 5.

Step 1: In standard form, the remainder must be from 0 to 499. Step 2: (500\times10=5000), so (5001=5000+1). Step 3: A negative remainder or a remainder greater than 500 does not make the standard Euclidean form.

Step 1: Check the squares of the given remainders. Step 2: (1^2) and (9^2) leave remainder 1, while (3^2) and (7^2) leave remainder 9. Step 3: When listing possible remainders, count repeated results only once.

Step 1: For eleven times the number, the remainder part is (11\times8=88). Step 2: (88=39\times2+10), so the final remainder is 10. Step 3: After multiplication, divide the result again by the divisor.

Step 1: The remainder of (a) is 7. Step 2: The remainder of (a^2+5) comes from (7^2+5=54). Step 3: (54=12\times4+6), so the final remainder is 6.

Step 1: In (4a+3b), the remainder part is (4\times9+3\times13=75). Step 2: (75=17\times4+7), so the remainder is 7. Step 3: In multi-term expressions, handle each term’s remainder separately.

Step 1: The original remainder is 57. Step 2: 175 leaves remainder 1 when divided by 58. Step 3: (57+1=58), so the final remainder is 0.

Step 1: Replace (n) by its remainder 4. Step 2: The remainder of (n^2+n+1) comes from (4^2+4+1=21). Step 3: Dividing 21 by 5 gives remainder 1.

Step 1: The remainder of (a) is 31. Step 2: For (8a), check (8\times31=248). Step 3: (248=42\times5+38), so the remainder is 38.

Step 1: The remainder must satisfy (0\le r<97). Step 2: 97 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 13, the remainder of (n+1) is 0. Step 3: When (n+1) is divisible by 14, its square is also divisible by 14.

Step 1: Add the remainders: (29+31+6=66). Step 2: 66 is exactly divisible by 33. Step 3: Therefore, the final remainder is 0; adding only the remainders is a quick method for multi-term expressions.