Euclid’s Division Lemma

Step 1: (4\times7=28), the nearest smaller multiple of (4) to (29). Step 2: (29-28=1), so the correct form is (29=4\times7+1). Step 3: Do not allow the remainder to be negative or equal to (4).

Step 1: The remainder is always less than the divisor. Step 2: The greatest integer less than (11) is (10). Step 3: The greatest possible remainder is always one less than the divisor.

Step 1: Remainder (0) means nothing is left after division. Step 2: Therefore, the dividend is exactly divisible by the divisor. Step 3: To identify exact divisibility, check the remainder.

Step 1: (10\times7=70) and (76-70=6). Step 2: The remainder (6) is less than (10), so the form is correct. Step 3: When dividing by (10), the last digit often helps find the remainder.

Step 1: Here the divisor is (6). Step 2: The remainder can be from (0) to (5), because it must be less than (6). Step 3: Include (0) in the list of possible remainders.

Step 1: (5\times3=15). Step 2: (18-15=3), so the remainder is (3). Step 3: Choose a multiple that does not exceed the dividend.

Step 1: In (a=bq+r), (a) is the dividend and (b) is the divisor. Step 2: (q) represents the quotient and (r) represents the remainder. Step 3: Remembering the meaning of symbols helps solve short questions quickly.

Step 1: The lemma is written as (a=bq+r). Step 2: Here (r) is the part left after division, so it is the remainder. Step 3: Identify (q) and (r) separately.

Step 1: (9\times4=36). Step 2: (40-36=4), so the remainder is (4). Step 3: In mental calculation, first take the nearest smaller multiple.

Step 1: In (a=bq+r), (b) is the divisor. Step 2: In the given form, (12) is multiplied by (q), so it is the divisor. Step 3: The number multiplying the quotient is usually the divisor.

Step 1: In (a=bq+r), (q) is the quotient. Step 2: In (12\times4), the number (4) is in the place of the quotient. Step 3: Match the whole form to identify divisor and quotient.

Step 1: To find the invalid form, check the range of the remainder. Step 2: In (23=5\times3+8), the remainder (8) is greater than the divisor (5). Step 3: Even if the equality is numerically true, check the remainder condition.

Step 1: When the divisor is (3), possible remainders are (0,1,2). Step 2: Remainder (3) equals the divisor, so it is not possible. Step 3: Treat a remainder equal to the divisor as incorrect.

Step 1: (10\times10=100). Step 2: (101=10\times10+1), so the quotient is (10). Step 3: The quotient is decided by the nearest smaller multiple.

Step 1: (6\times6=36). Step 2: (37-36=1), and (1<6), so the form is correct. Step 3: Wrong options can be removed quickly by checking the remainder range.

Step 1: The condition becomes (0\le r<1). Step 2: Only (0) satisfies this range, so the remainder is (0). Step 3: Use the inequality range to solve such questions.

Step 1: With divisor (2), the remainder must satisfy (0\le r<2). Step 2: Hence (r=0) or (r=1). Step 3: This idea also helps in understanding even and odd numbers.

Step 1: When a number is divided by (2), the remainder can be (0) or (1). Step 2: An even number has remainder (0), so its form is (2q). Step 3: Remember (2q) and (2q+1) for even and odd number questions.

Step 1: Dividing by (2) gives remainder (0) or (1). Step 2: An odd number has remainder (1), so its form is (2q+1). Step 3: The remainder helps identify the type of number.

Step 1: (8\times9=72). Step 2: (73-72=1), so the remainder is (1) and quotient is (9). Step 3: Choose the form where the remainder is less than the divisor.