Euclid’s Division Lemma

Step 1: Find the nearest lower multiple of 84 below 1025. Step 2: (84\times12=1008), so the remainder is (1025-1008=17). Step 3: The final remainder must be less than 84 for the form to be valid.

Step 1: The number is of the form (31\times27+r), where (0\le r<31). Step 2: The greatest remainder is 30, so the number is (837+30=867). Step 3: For the greatest value, take the remainder as one less than the divisor.

Step 1: The number is (46\times19+r). Step 2: For the least value, take remainder 0, so the number is (46\times19=874). Step 3: For minimum value questions, taking remainder zero is the safest method.

Step 1: A valid remainder must be between 0 and 57. Step 2: (58\times12=696), so (703=696+7), and 7 is valid. Step 3: Along with calculation, also check the range of the remainder.

Step 1: The remainder of (x) is 14. Step 2: Adding 16 gives total remainder (14+16=30), which is exactly divisible by 15. Step 3: After addition, reduce the remainder again by the divisor.

Step 1: Let (n=17q+6). Step 2: (4n+5=68q+24+5=68q+29=17(4q+1)+12). Step 3: In a linear expression, first multiply the remainder and then reduce by the divisor.

Step 1: The square remainder comes from dividing (7^2=49) by 9. Step 2: (49=9\times5+4), so the remainder is 4. Step 3: In square questions, squaring only the remainder is faster than using the whole number.

Step 1: On division by 6, possible remainders are 0, 1, 2, 3, 4, and 5. Step 2: So the forms are from (6q) to (6q+5). Step 3: Include remainder 0 and do not include 6 in the complete list.

Step 1: The remainder of (a) is 22, one less than 23. Step 2: Adding 1 gives (23q+23=23(q+1)), so the remainder is 0. Step 3: Adding 1 to a remainder one less than the divisor takes the number to the next multiple.

Step 1: Write (m=12q+5). Step 2: (m-19=12q-14=12(q-2)+10), so the remainder is 10. Step 3: If subtraction gives a negative remainder, add the divisor enough times to make it valid.

Step 1: The condition on the remainder is (0\le r<b). Step 2: If (b=28), the greatest possible value of (r) is 27. Step 3: The remainder can never be equal to the divisor.

Step 1: Let the number be (25q+24). Step 2: For twice the number, the remainder part is (2\times24=48), and (48=25+23). Step 3: After multiplication, reduce the remainder below 25.

Step 1: Find the nearest lower multiple of 36 below 589. Step 2: (36\times16=576), so the remainder is (589-576=13). Step 3: For larger numbers, use the nearest lower multiple.

Step 1: The cube remainder comes from (3^3=27). Step 2: (27=8\times3+3), so the remainder is 3. Step 3: In power questions, keep the calculation small by using the remainder.

Step 1: In Euclid’s lemma, the remainder starts from 0. Step 2: When divided by 10, possible remainders are 0 through 9. Step 3: Do not include the divisor itself in the list of remainders.

Step 1: The remainder of (p) is 8. Step 2: For (3p-7), the remainder part is (3\times8-7=17), and (17=11+6). Step 3: In a linear expression, always reduce the final remainder below the divisor.

Step 1: The two remainders are 4 and 5. Step 2: The sum has remainder (4+5=9), and (9=7+2), so the remainder is 2. Step 3: If the sum of remainders is greater than the divisor, reduce it again.

Step 1: For multiplication, multiply the remainders 10 and 9. Step 2: (10\times9=90), and (90=13\times6+12). Step 3: In products, using remainders instead of whole numbers saves time.

Step 1: For the cube, consider (2^3=8). Step 2: (8=5\times1+3), so the cube leaves remainder 3. Step 3: In powers, first raise the small remainder to the power.

Step 1: The possible remainders of a number are 0, 1, 2, and 3. Step 2: The square remainders are respectively 0, 1, 0, and 1. Step 3: A square divided by 4 never leaves remainder 2 or 3.