Euclid’s Division Lemma

Step 1: On division by (4), the possible remainders are (0,1,2,3). Step 2: Hence the forms are (4q,4q+1,4q+2,4q+3). Step 3: These forms are very useful in questions related to even and odd numbers.

Step 1: (19 \times 8=152) and (19 \times 9=171). Step 2: (152) is less than (158), so (q=8) and (r=158-152=6). Step 3: If the next multiple exceeds the number, take the previous multiple.

Step 1: In the form (a=bq+r), (a) is the dividend. Step 2: (b) is the number by which division is done, so it is the divisor. Step 3: Keep the meaning of symbols clear to avoid calculation errors.

Step 1: The remainder should not be negative and must be less than (10). Step 2: (10 \times 8=80) and the remainder is (9), so (89=10 \times 8+9). Step 3: A form with a negative remainder is not the standard Euclidean form.

Step 1: Substitute the values in (a=bq+r). Step 2: (a=9 \times 5+7=45+7=52). Step 3: You can check the result by dividing it again by (9).

Step 1: Find the nearest multiple of (24) less than (200). Step 2: (24 \times 8=192) and (200-192=8), so the remainder is (8). Step 3: Practice finding nearby multiples for faster calculation.

Step 1: Compare with the Euclidean form (a=bq+r). Step 2: Here the divisor is (17) and the remainder is (13), since (13<17). Step 3: Always check the range of the remainder while comparing.

Step 1: The remainder must be smaller than (11). Step 2: (15=11+4), so (11q+15=11(q+1)+4). Step 3: If the remainder is greater than the divisor, divide it again and rewrite the form.

Step 1: Remainder (0) means the number is exactly divisible by (6). Step 2: So the number is (6q+0), that is (6q). Step 3: Forms with zero remainder help identify multiples.

Step 1: (32 \times 8=256). Step 2: (257-256=1), so (257=32 \times 8+1). Step 3: In the correct form, the remainder must be smaller than (32) and not negative.

Step 1: This lemma is used for two positive integers. Step 2: In it, (a) and (b) are positive integers and (b \ne 0). Step 3: While reading the question, pay attention to the type of numbers involved.

Step 1: (30 \times 12=360). Step 2: (365-360=5), so the quotient is (12) and the remainder is (5). Step 3: Keeping the remainder smaller than the divisor decides whether the answer is valid.

Step 1: Compare it with (a=bq+r). Step 2: In (a=5q+4), the divisor is (5) and the remainder is (4). Step 3: Since (4<5), this form is already correct.

Step 1: On division by (2), the remainder can only be (0) or (1). Step 2: So the forms are (2q) or (2q+1). Step 3: This is the basis for identifying even and odd numbers.

Step 1: (20 \times 7=140). Step 2: (148-140=8), so (r=8). Step 3: If an option gives a remainder equal to or greater than (20), reject it immediately.

Step 1: The remainder cannot be (14) because it equals the divisor. Step 2: (14q+14=14(q+1)+0), so the correct remainder is (0). Step 3: If the remainder equals the divisor, increase the quotient by one.

Step 1: According to Euclid’s division lemma, (a=bq+r). Step 2: Here the divisor is (11) and the remainder is (3), so the form is (11q+3). Step 3: While writing a general form, multiply the divisor by (q).

Step 1: (13 \times 7=91). Step 2: (91-91=0), so the remainder is (0). Step 3: When division is exact, the remainder is always (0).

Step 1: Substitute in (a=bq+r): (73=7b+3). Step 2: (70=7b), so (b=10). Step 3: To find an unknown divisor, subtract the remainder first.

Step 1: The condition for the remainder is (0 \le r < b). Step 2: Here (r=15) and (b=16), so it is valid because (15<16). Step 3: The greatest possible remainder is (b-1).