Euclid’s Division Lemma

Step 1: Find the nearest lower multiple of 156 below 2547. Step 2: (156\times16=2496), so the remainder is (2547-2496=51). Step 3: In a valid Euclidean form, the remainder must be smaller than the divisor.

Step 1: The number is of the form (64\times37+r), where (0\le r<64). Step 2: The greatest remainder is 63, so the number is (2368+63=2431). Step 3: For the greatest value, take the remainder as one less than the divisor.

Step 1: The number is (91\times28+r). Step 2: For the least value, the remainder is 0, so the number is (91\times28=2548). Step 3: In minimum value questions, taking remainder zero gives the answer quickly.

Step 1: In Euclidean form, the remainder is non-negative and smaller than 247. Step 2: (247\times12=2964), so (3199=2964+235). Step 3: A negative remainder or a remainder bigger than the divisor is not the standard form.

Step 1: The remainder of (x) is 41. Step 2: 89 leaves remainder 3 when divided by 43, so the total remainder is (41+3=44). Step 3: Since (44=43+1), the final remainder is 1.

Step 1: Replace (n) by its remainder 17. Step 2: The remainder of (7n+13) comes from (7\times17+13=132). Step 3: (132=29\times4+16), so the correct remainder is 16.

Step 1: The square remainder comes from dividing (11^2=121) by 13. Step 2: (121=13\times9+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 11, remainders can be from 0 to 10. Step 2: Therefore, all forms are from (11q) to (11q+10). Step 3: Include remainder 0, but do not include 11.

Step 1: The remainder of (a) is 70, one less than 71. Step 2: Adding 1 gives (71q+71=71(q+1)). Step 3: The number becomes exactly divisible by 71, so the remainder is 0.

Step 1: Write (m=31q+6). Step 2: (m-68=31q-62=31(q-2)), so the remainder is 0. Step 3: In subtraction, check the final form as divisor times quotient plus remainder.

Step 1: The remainder condition is (0\le r<108). Step 2: The greatest integer smaller than 108 is 107. Step 3: Remember in exams that the remainder is never equal to the divisor.

Step 1: For five times the number, the remainder part is (5\times33=165). Step 2: (165=34\times4+29), so the final remainder is 29. Step 3: After multiplication, always reduce the result below the divisor.

Step 1: Find the nearest lower multiple of 132 below 4217. Step 2: (132\times31=4092), so the remainder is (4217-4092=125). Step 3: For large numbers, the nearest lower multiple method is useful.

Step 1: For the cube, consider (5^3=125). Step 2: (125=12\times10+5), so the remainder is 5. Step 3: In power questions, reduce remainders along the way to keep calculation easy.

Step 1: In Euclid’s lemma, (0\le r<b). Step 2: Here (b=18), so the remainder can be from 0 to 17. Step 3: Include 0 in the list of remainders and do not include the divisor.

Step 1: The remainder of (p) is 14. Step 2: The remainder of (5p-19) comes from (5\times14-19=51). Step 3: Since (51=17\times3), the final remainder is 0.

Step 1: The two remainders are 16 and 15. Step 2: The sum remainder comes from (16+15=31). Step 3: Since (31=19+12), the final remainder is 12.

Step 1: For multiplication, multiply the remainders 19 and 21. Step 2: (19\times21=399), and (399=23\times17+8). Step 3: In product questions, the final answer must be the remainder, not the full product.

Step 1: For the cube, consider (4^3=64). Step 2: (64=9\times7+1), so the remainder is 1. Step 3: In powers, working with the small remainder avoids large calculations.

Step 1: On division by 7, remainders can be from 0 to 6. Step 2: Their square remainders are 0, 1, 4, 2, 2, 4, and 1. Step 3: So the square remainder can only be 0, 1, 2, or 4.