Euclid’s Division Lemma

Step 1: Dividing (56) by (7) gives (7 \times 8=56). Step 2: Nothing is left so the remainder is (0). Step 3: When the dividend is a multiple of the divisor, the remainder is always (0).

Step 1: The nearest multiple of (5) below (38) is (35). Step 2: (38-35=3), so (q=7) and (r=3). Step 3: In the correct answer, the remainder must be less than (5).

Step 1: A remainder may start from (0). Step 2: The remainder is always less than the divisor, so division by (6) gives remainders from (0) to (5). Step 3: Do not include the divisor itself while listing possible remainders.

Step 1: Use the form (a=bq+r). Step 2: (a=9 \times 4+2=36+2=38). Step 3: In such questions, first multiply the divisor and quotient, then add the remainder.

Step 1: In Euclid’s division lemma, the remainder is not negative. Step 2: The remainder is smaller than the divisor, so (0 \le r < b) is correct. Step 3: Avoid the common mistake of allowing (r=b).

Step 1: (10 \times 7=70). Step 2: (72-70=2), so (72=10 \times 7+2) is correct. Step 3: Along with equality, the remainder must also be less than (10).

Step 1: When divided by (2), the remainder can be (0) or (1). Step 2: So the number has the form (2q) or (2q+1). Step 3: This idea is the base for understanding even and odd numbers.

Step 1: (8 \times 5=40) and (8 \times 6=48). Step 2: Since (48) is greater, take (40) and get (47-40=7). Step 3: If the next multiple is greater, use the previous multiple.

Step 1: The condition for the remainder is (0 \le r < b). Step 2: Here (b=11), so (r=11) is not valid. Step 3: The remainder is never equal to the divisor.

Step 1: (12 \times 7=84) and (12 \times 8=96). Step 2: Since (96) is greater than (95), the quotient is (7). Step 3: The quotient is the greatest integer whose product with the divisor does not exceed the dividend.

Step 1: (12 \times 7=84). Step 2: (95-84=11), so the remainder is (11). Step 3: The answer is valid because (11<12).

Step 1: Compare with (a=bq+r). Step 2: Here the divisor is (13) and the remainder is (5). Step 3: Since (5<13), the form is already correct.

Step 1: On division by (4), possible remainders are (0,1,2,3). Step 2: In (4q+4), the remainder is (4), which equals the divisor. Step 3: Such a form should be written as (4(q+1)).

Step 1: (20 \times 6=120). Step 2: (123-120=3), so (123=20 \times 6+3). Step 3: A negative remainder or a remainder greater than (20) is not correct.

Step 1: In (a=bq+r), (a) is the dividend and (b) is the divisor. Step 2: (q) represents the quotient. Step 3: Remembering the names of symbols makes the formula easier to use.

Step 1: The greatest remainder is one less than the divisor. Step 2: On division by (15), the greatest remainder is (15-1=14). Step 3: When greatest remainder is asked, use (b-1).

Step 1: The dividend (17) is smaller than the divisor (23). Step 2: So the quotient is (0) and the remainder remains (17). Step 3: When the dividend is smaller than the divisor, remember to take (q=0).

Step 1: (13 \times 11=143). Step 2: (144-143=1), so the remainder is (1). Step 3: Finding the nearest smaller multiple saves time.

Step 1: The remainder must be less than (9). Step 2: (12=9+3), so (9q+12=9(q+1)+3). Step 3: If the leftover part is greater than the divisor, divide it again.

Step 1: On division by (3), possible remainders are (0,1,2). Step 2: In (3q+3), the remainder is (3), which equals the divisor. Step 3: It should be written as (3(q+1)).