Euclid’s Division Lemma

Step 1: The remainder condition is (0\le r<52). Step 2: The greatest integer smaller than 52 is 51, so it is the greatest possible remainder. Step 3: A remainder can never be equal to the divisor.

Step 1: For three times the number, the remainder part is (3\times27=81). Step 2: (81=28\times2+25), so the final remainder is 25. Step 3: After multiplication, reduce the remainder below the divisor.

Step 1: Find the nearest lower multiple of 41 below 875. Step 2: (41\times21=861), so the remainder is (875-861=14). Step 3: The nearest lower multiple method saves time with larger numbers.

Step 1: For the cube, take (3^3=27). Step 2: Dividing 27 by 10 gives remainder 7. Step 3: In power questions, use the smaller remainder instead of the full number.

Step 1: In Euclid’s lemma, (0\le r<b). Step 2: Here (b=14), so the remainder can be from 0 to 13. Step 3: Include 0 in the list of remainders and exclude the divisor itself.

Step 1: The remainder of (p) is 11. Step 2: The remainder of (4p-9) comes from (4\times11-9=35), and (35=13\times2+9). Step 3: Always reduce the final remainder below the divisor.

Step 1: The two remainders are 6 and 7. Step 2: The sum remainder comes from (6+7=13), and (13=8+5). Step 3: If the sum of remainders is greater than the divisor, reduce it again.

Step 1: For multiplication, multiply the remainders 12 and 15. Step 2: (12\times15=180), and (180=17\times10+10). Step 3: In product questions, multiply the remainders and then find the final remainder.

Step 1: For the cube, consider (5^3=125). Step 2: (125=7\times17+6), so the remainder is 6. Step 3: In higher powers, reduce remainders along the way to keep calculation easy.

Step 1: The possible remainders of a number are 0, 1, 2, 3, and 4. Step 2: Their square remainders are 0, 1, 4, 4, and 1 respectively. Step 3: A square divided by 5 never leaves remainder 2 or 3.

Step 1: The remainder of (N) is 17. Step 2: (49) leaves remainder 5 on division by 22, so (17+5=22), which gives remainder 0. Step 3: After reducing the added number, reduce the final sum again by the divisor.

Step 1: When the divisor is 24, the remainder can be from 0 to 23. Step 2: 24 is equal to the divisor, so it cannot be a remainder. Step 3: In remainder questions, carefully check any option equal to the divisor.

Step 1: (67\times25=1675). Step 2: (1682-1675=7), so the remainder is 7. Step 3: While dividing, choose a multiple that does not exceed the given number.

Step 1: The remainder of (t) is 13. Step 2: The remainder part of (3t+4) is (3\times13+4=43), and (43=18\times2+7). Step 3: The final remainder must be reduced below 18.

Step 1: For seven times the number, the remainder part is (7\times6=42). Step 2: (42=32+10), so the final remainder is 10. Step 3: After multiplication, reduce the result again by the divisor.

Step 1: Any integer divided by 5 has one of the forms (5q), (5q+1), (5q+2), (5q+3), or (5q+4). Step 2: Five consecutive integers cover all five remainders. Step 3: The number with remainder 0 is divisible by 5.

Step 1: The possible remainders are 1 and 5. Step 2: (1^2=1) and (5^2=25=6\times4+1), so the remainder is 1 in both cases. Step 3: In such forms, work only with the remainder, not the whole number.

Step 1: (a-48=37q+29-48=37q-19). Step 2: This can be written as (37(q-1)+18), so the remainder is 18. Step 3: Add the divisor once to make a negative remainder valid.

Step 1: For the square, take (11^2=121). Step 2: (121=15\times8+1), so the remainder is 1. Step 3: Do not write 121 as the final answer because a remainder must be smaller than the divisor.

Step 1: The remainder of (x) is 7. Step 2: The remainder of (x^2+x) comes from (7^2+7=56), and 56 is divisible by 8. Step 3: Substituting the remainder in the expression gives a quick solution.