Euclid’s Division Lemma

Step 1: A number can be (3q), (3q+1), or (3q+2). Step 2: The square remainders are 0, 1, and the remainder of (2^2=4), which is 1. Step 3: For squares modulo 3, remainder 2 never appears.

Step 1: On division by 5, the remainder can be 0, 1, 2, 3, or 4. Step 2: 6 is greater than divisor 5, so it cannot be a remainder. Step 3: First eliminate options equal to or greater than the divisor.

Step 1: The remainder of (N) is 11. Step 2: The remainder of 25 on division by 12 is 1, so total remainder (11+1=12), which becomes 0. Step 3: In addition, add the remainders and then reduce by the divisor.

Step 1: Let (y=9q+2). Step 2: (5y+4=45q+10+4=45q+14=9(5q+1)+5). Step 3: For a linear expression, replace the number by its remainder and simplify.

Step 1: The number is (14q+13). Step 2: Subtracting 2 gives (14q+11), so the remainder is 11. Step 3: In subtraction, if the remainder does not become negative, subtract directly.

Step 1: Let (z=8q+1). Step 2: Use the remainder in the power: (1^3=1), so the remainder on division by 8 is still 1. Step 3: For powers, you do not need to expand the whole expression.

Step 1: Let the number be (16q+15). Step 2: The square remainder is obtained from (15^2=225) divided by 16. (225=16\times14+1). Step 3: The square of a remainder (b-1) often gives remainder 1.

Step 1: On division by 7, remainders can be from 0 to 6. Step 2: So the forms are from (7q) to (7q+6). Step 3: Include remainder 0 and do not include remainder 7 in the complete list.

Step 1: The remainder of (a) is 19. Step 2: Adding 1 gives (20q+20=20(q+1)), so the remainder is 0. Step 3: Adding 1 to a remainder one less than the divisor gives the next multiple.

Step 1: For the cube, work with the remainder (2^3=8). Step 2: (8=6\times1+2), so the cube leaves remainder 2. Step 3: In powers, keep the calculation small by using the remainder.

Step 1: The remainders of the two numbers are 5 and 7. Step 2: Their sum gives remainder (5+7=12), which is less than 18. Step 3: For a sum, adding the remainders first is easier.

Step 1: Add the remainders 14 and 11. Step 2: (14+11=25), and (25=18+7), so the remainder is 7. Step 3: If the sum of remainders exceeds the divisor, reduce it again.

Step 1: In multiplication, multiply the remainders. Step 2: (3\times4=12), and 12 leaves remainder 2 on division by 10. Step 3: For products, multiply the remainders instead of the whole numbers.

Step 1: The number is (13q+12). Step 2: For three times the number, the remainder part is (3\times12=36), and (36=13\times2+10). Step 3: After multiplication, reduce the remainder below the divisor.

Step 1: The possible remainders are 1 and 3. Step 2: (1^2=1) and (3^2=9=4\times2+1), so the remainder is 1 in both cases. Step 3: The square of an odd number leaves remainder 1 when divided by 4.

Step 1: The square remainder comes from dividing (4^2=16) by 5. Step 2: (16=5\times3+1), so the remainder is 1. Step 3: A remainder one less than the divisor often gives square remainder 1.

Step 1: Write (a=7q+5). Step 2: (a-9=7q-4=7(q-1)+3), so the remainder is 3. Step 3: If subtraction gives a negative remainder, add the divisor to make it valid.

Step 1: The number is (9q+8). Step 2: Adding 4 gives (9q+12=9(q+1)+3), so the remainder is 3. Step 3: Reduce the new remainder below the divisor by subtracting 9.

Step 1: Find the nearest lower multiple of 45 below 728. Step 2: (45\times16=720), so (728-720=8). Step 3: The nearest-multiple method saves time in large divisions.

Step 1: The number is (12q+7). Step 2: Adding 29 gives total remainder (7+29=36), and 36 is exactly divisible by 12. Step 3: You may add the given number directly, then find the final remainder.