If (b=1), what will be the remainder when any positive integer (a) is divided by (1)?
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.
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SubjectsMathematics
यूक्लिड का विभाजन प्रमेय
In Class 10 Mathematics, under the chapter Real Numbers, Euclid’s Division Lemma introduces the relationship a = bq + r, where a and b are positive integers, q is the quotient, and the remainder r satisfies 0 ≤ r < b. Students learn how repeated division forms Euclid’s division algorithm and use it to find the highest common factor (HCF) of two numbers. The topic also strengthens understanding of divisibility, quotients, remainders, and the logical steps used in number-theory proofs.
TOPIC PRACTICE
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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.
Step 1: (7\times12=84). Step 2: The division is exact, so the quotient is (12) and the remainder is (0). Step 3: Strong multiplication tables make such questions quick.
Step 1: Compare with the form (a=bq+r). Step 2: Here the divisor is (15) and the remainder is (14), which is less than (15). Step 3: When the form is already given, the remainder can be read directly.
Step 1: Here the divisor appears to be (7) and the remainder (7). Step 2: The remainder must be less than the divisor, not equal to it. Step 3: If the remainder equals the divisor, it should be carried into the quotient.
Step 1: In (7q+7), the extra (7) can be treated as (7\times1). Step 2: So (7q+7=7(q+1)+0), where the remainder is (0). Step 3: If the remainder equals the divisor, add (1) to the quotient.
Step 1: (5\times13=65). Step 2: (66-65=1), so the remainder is (1). Step 3: For division by (5), the last digit can also help.
Step 1: In (a=bq+r), (a) is the dividend. Step 2: In the given form, (99) is on the left side, so the dividend is (99). Step 3: The dividend is the number being divided.
Step 1: (16\times7=112) and (16\times8=128), which is greater than (120). Step 2: (120-112=8), so the remainder is (8). Step 3: Never choose a multiple greater than the dividend.
Step 1: In (a=4q+r), the divisor is (4). Step 2: The remainder can be (0,1,2,3), but not (4). Step 3: The upper limit of the remainder is one less than the divisor.
Step 1: In (a=9q+0), the remainder is (0). Step 2: A zero remainder means the number is exactly divisible by (9). Step 3: Treat zero remainder as a sign of exact division.
Step 1: The divisor cannot be zero in division. Step 2: In the lemma, (b) is taken as a positive integer. Step 3: Division by zero is not valid, so avoid such options.
Step 1: The condition is (0\le r<10). Step 2: So the smallest possible value of (r) is (0). Step 3: The remainder range always starts from (0).
Step 1: The divisor is (10), so the remainder must be less than (10). Step 2: The greatest integer less than (10) is (9). Step 3: To get the greatest remainder, subtract (1) from the divisor.
Step 1: (11\times4=44). Step 2: (45-44=1), which is less than (11). Step 3: A correct Euclidean form always has the remainder in the valid range.
Step 1: The divisor is (7), so (7q) represents a multiple of it. Step 2: Adding remainder (6) gives (7q+6). Step 3: In such questions, put the divisor in the multiple part.
Step 1: (12q) is exactly divisible by (12). Step 2: The leftover part is (5), and it is less than (12), so it is the remainder. Step 3: In (bq+r), (bq) is the multiple and (r) is the remainder.
Step 1: (9\times11=99). Step 2: (100-99=1), so the quotient is (11) and the remainder is (1). Step 3: Do not accept options with an oversized remainder in Euclid’s lemma.
Step 1: With divisor (3), the remainders can be (0,1,2). Step 2: So there are three different possible remainders. Step 3: For a divisor (b), the number of possible remainders is (b).
Step 1: (4\times6=24) and (4\times7=28). Step 2: Since (28) is greater than (27), the quotient is (6). Step 3: For the quotient, choose a multiple that does not exceed the dividend.
Step 1: We can write (27=4\times6+3). Step 2: So the remainder is (3), and it is less than (4). Step 3: After finding the quotient, subtract to get the remainder.
Step 1: This lemma explains the basic structure of division. Step 2: It helps write the dividend, divisor, quotient, and remainder in the form (a=bq+r). Step 3: In the chapter on real numbers, it becomes a base for later methods.
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