Euclid’s Division Lemma

Step 1: Find the nearest lower multiple of 238 below 4961. Step 2: (238\times20=4760), so the remainder is (4961-4760=201). Step 3: Since the remainder is smaller than 238, this is the valid Euclidean form.

Step 1: The number is of the form (87\times46+r), where (0\le r<87). Step 2: The greatest remainder is 86, so the number is (4002+86=4088). Step 3: For the greatest value, the remainder is always one less than the divisor.

Step 1: The number is (143\times23+r). Step 2: For the least value, the remainder is 0, so the number is (143\times23=3289). Step 3: In least-value questions, taking the remainder as zero is the clearest method.

Step 1: In standard form, the remainder must be from 0 to 390. Step 2: (391\times15=5865), so (6127=5865+262). Step 3: A negative remainder or a remainder greater than 391 does not make the correct Euclidean form.

Step 1: The remainder of (x) is 64. Step 2: 140 leaves remainder 6 when divided by 67, so the total remainder is (64+6=70). Step 3: Since (70=67+3), the final remainder is 3.

Step 1: Replace (n) by its remainder 31. Step 2: The remainder of (9n+22) comes from (9\times31+22=301). Step 3: Since (301=43\times7+0), the remainder is 0.

Step 1: The square remainder comes from dividing (16^2=256) by 19. Step 2: (256=19\times13+9), so the remainder is 9. Step 3: In square questions, square only the remainder instead of the full number.

Step 1: On division by 17, remainders can be from 0 to 16. Step 2: Therefore, all forms are from (17q) to (17q+16). Step 3: Include zero remainder, but do not include 17.

Step 1: The remainder of (a) is 96, one less than 97. Step 2: Adding 1 gives (97q+97=97(q+1)). Step 3: The number becomes exactly divisible by 97, so the remainder is 0.

Step 1: Write (m=53q+12). Step 2: (m-118=53q+12-118=53q-106=53(q-2)). Step 3: It is exactly divisible by 53, so the remainder is 0.

Step 1: The remainder condition is (0\le r<144). Step 2: The greatest integer smaller than 144 is 143. Step 3: The remainder is never equal to the divisor.

Step 1: For nine times the number, the remainder part is (9\times58=522). Step 2: (522=59\times8+50), so the remainder is 50. Step 3: After multiplication, always reduce the result below the divisor.

Step 1: Find the nearest lower multiple of 359 below 9264. Step 2: (359\times25=8975), so the remainder is (9264-8975=289). Step 3: With large numbers, the nearest lower multiple method saves time.

Step 1: For the cube, check (11^3=1331). Step 2: (1331=18\times73+17), so the remainder is 17. Step 3: In power questions, reduce remainders repeatedly to keep the calculation simple.

Step 1: In Euclid’s lemma, (0\le r<b). Step 2: Here (b=32), so the remainder can be from 0 to 31. Step 3: Include 0 in the list of remainders and do not include 32.

Step 1: The remainder of (p) is 19. Step 2: The remainder of (7p-18) comes from (7\times19-18=115). Step 3: Since (115=23\times5), the final remainder is 0.

Step 1: The two remainders are 25 and 28. Step 2: The sum remainder comes from (25+28=53). Step 3: Since (53=31+22), the final remainder is 22.

Step 1: For multiplication, multiply the remainders 32 and 35. Step 2: (32\times35=1120), and (1120=37\times30+10). Step 3: In product questions, the answer is the final remainder, not the full product.

Step 1: For the cube, check (6^3=216). Step 2: (216=13\times16+8), so the remainder is 8. Step 3: In powers, working with small remainders avoids large calculations.

Step 1: On division by 11, remainders can be from 0 to 10. Step 2: Their distinct square remainders are 0, 1, 3, 4, 5, and 9. Step 3: In square-remainder questions, making a short list of possible remainders is useful.