Euclid’s Division Lemma

Step 1: On division by 9, possible remainders are from 0 to 8. Step 2: Nine consecutive integers cover all these remainders once. Step 3: The number with remainder 0 is divisible by 9.

Step 1: The remainders in the two forms are 3 and 7. Step 2: (3^2=9) and (7^2=49), and both leave remainder 9 when divided by 10. Step 3: In form-based questions, work only with the remainder.

Step 1: (a-119=67q+52-119=67q-67). Step 2: This can be written as (67(q-1)+0). Step 3: The number is exactly divisible by 67, so the remainder is 0.

Step 1: For the square, take (21^2=441). Step 2: (441=26\times16+25), so the remainder should be 25. Step 3: The final answer must always be less than 26.

Step 1: The remainder of (x) is 11. Step 2: The remainder of (x^2+x) comes from (11^2+11=132). Step 3: Since 132 is exactly divisible by 12, the remainder is 0.

Step 1: (137\times29=3973) and (137\times30=4110). Step 2: The nearest lower multiple below 4097 is 3973, so the remainder is (4097-3973=124). Step 3: Do not accept a negative remainder or a remainder greater than 137.

Step 1: Number (=) divisor (\times) quotient (+) remainder. Step 2: (96\times31+95=2976+95=3071). Step 3: The remainder 95 is less than divisor 96, so the form is valid.

Step 1: For the difference, the remainder is (8-19=-11). Step 2: Add 35 to make it a valid remainder, giving 24. Step 3: In subtraction, add the divisor when the remainder becomes negative.

Step 1: On division by 10, possible remainders are from 0 to 9. Step 2: Ten consecutive integers cover all these remainders once. Step 3: The number with remainder 0 is divisible by 10.

Step 1: The original remainder is 61. Step 2: 187 leaves remainder 1 when divided by 62. Step 3: The total remainder is (61+1=62), so the final remainder is 0.

Step 1: The remainder of (u) is 11. Step 2: The remainder of (u^2-u) comes from (11^2-11=110). Step 3: (110=18\times6+2), so the remainder is 2.

Step 1: (10^2=100), and 100 leaves remainder 15 when divided by 17. Step 2: For (10^4), check (15^2=225). Step 3: (225=17\times13+4), so the remainder is 4.

Step 1: The original remainder is 29. Step 2: 43 leaves remainder 7 when divided by 36. Step 3: (29+7=36), so the new remainder is 0.

Step 1: The remainders of (a) and (b) are 13 and 18. Step 2: The remainder of (ab+a) comes from (13\times18+13=247). Step 3: (247=20\times12+7), so the final remainder is 7.

Step 1: In standard form, the remainder must be from 0 to 699. Step 2: (700\times10=7000), so (7001=7000+1). Step 3: A negative remainder or a remainder greater than 700 does not make the standard Euclidean form.

Step 1: Check the squares of the given remainders. Step 2: (1^2), (5^2), (7^2), and (11^2) all leave remainder 1 when divided by 12. Step 3: Checking the squares of possible remainders separately is a safe method.

Step 1: For thirteen times the number, the remainder part is (13\times11=143). Step 2: (143=45\times3+8), so the final remainder is 8. Step 3: After multiplication, divide the result again by the divisor.

Step 1: The remainder of (a) is 9. Step 2: The remainder of (a^2+7) comes from (9^2+7=88). Step 3: (88=16\times5+8), so the final remainder is 8.

Step 1: In (5a+4b), the remainder part is (5\times11+4\times16=119). Step 2: (119=19\times6+5), so the remainder is 5. Step 3: In multi-term expressions, handle each term’s remainder separately.

Step 1: The original remainder is 73. Step 2: 223 leaves remainder 1 when divided by 74. Step 3: (73+1=74), so the final remainder is 0.

Step 1: Replace (n) by its remainder 5. Step 2: The remainder of (n^2+n+1) comes from (5^2+5+1=31). Step 3: Dividing 31 by 6 gives remainder 1.

Step 1: The remainder of (a) is 37. Step 2: For (10a), check (10\times37=370). Step 3: (370=48\times7+34), so the remainder is 34.

Step 1: The remainder must satisfy (0\le r<109). Step 2: 109 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 21, the remainder of (n+1) is 0. Step 3: When (n+1) is divisible by 22, its square is also divisible by 22.

Step 1: Add the remainders: (35+37+6=78). Step 2: 78 is exactly divisible by 39. Step 3: Therefore, the final remainder is 0; adding only the remainders is a quick method for multi-term expressions.