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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
Up to 25 questions from this page. Select your focus, then start.
Step 1: In (a=bq+r), (q) is the quotient. Step 2: The number multiplied with (47) is (13), so (13) is the quotient. Step 3: To identify the quotient, look at the multiplier written with the divisor.
Step 1: In (a=bq+r), the final added part is (r). Step 2: Here (7) is less than (47), so it is the remainder. Step 3: While identifying the remainder, also check its range.
Step 1: Substitute in (a=bq+r): (93=5b+8). Step 2: (85=5b), so (b=17). Step 3: To find the unknown divisor, subtract the remainder first.
Step 1: Substitute in (a=bq+r): (108=15q+3). Step 2: (105=15q), so (q=7). Step 3: To find the unknown quotient, subtract the remainder first.
Step 1: The Euclidean division form is (a=bq+r). Step 2: Here the divisor is (14) and the remainder is (9), so the form is (14q+9). Step 3: In a general form, multiply the divisor by (q) and add the remainder.
Step 1: On division by (9), remainders can be from (0) to (8). Step 2: In (9q+9), the remainder is (9), which equals the divisor. Step 3: It should be written correctly as (9(q+1)).
Step 1: (33 \times 30=990). Step 2: (1000-990=10), so the remainder is (10). Step 3: For a large dividend, finding a nearby multiple saves time.
Step 1: (33 \times 30=990) and (33 \times 31=1023). Step 2: Since (1023) is greater, the quotient is (30). Step 3: The quotient is the greatest integer whose product with the divisor does not exceed the dividend.
Step 1: (2a=2(5q+4)=10q+8). Step 2: (8=5+3), so (2a=5(2q+1)+3) and the remainder is (3). Step 3: In multiplication-based questions, first multiply the old remainder and then divide by the divisor.
Step 1: The remainder of (a) is (5), so for (3a), check (3 \times 5=15). Step 2: (15=7 \times 2+1), so the remainder is (1). Step 3: In such questions, work with the remainder instead of the whole number.
Step 1: Write (a=4q+3). Step 2: (2a=8q+6=4(2q+1)+2), so the remainder is (2). Step 3: After multiplication, do not forget to convert the remainder into the correct range.
Step 1: Let (a=6q+5). Step 2: (2a=12q+10=6(2q+1)+4), so the remainder is (4). Step 3: (10) cannot be the remainder because it is greater than (6).
Step 1: The remainder is always less than the divisor. Step 2: The greatest integer less than (25) is (24). Step 3: The greatest remainder can be found quickly using (b-1).
Step 1: A remainder may start from (0). Step 2: If a number is exactly divisible by (25), the remainder is (0). Step 3: The smallest possible remainder is always (0).
Step 1: (39) is smaller than (100), so the quotient is (0). Step 2: (39=100 \times 0+39), and (39<100), so the form is correct. Step 3: If the dividend is smaller, the remainder can be the dividend itself.
Step 1: (39 \times 2=78) and (39 \times 3=117). Step 2: Since (117) is greater, (q=2) and (r=100-78=22). Step 3: The remainder must be less than (39).
Step 1: Write (a=bq+(b-1)). Step 2: (a+1=bq+b=b(q+1)+0), so the remainder is (0). Step 3: Adding (1) to the greatest remainder brings the remainder back to (0).
Step 1: (a=bq) is exactly divisible by (b). Step 2: (a+1=bq+1), so the remainder is (1). Step 3: Just after an exactly divisible number, the remainder is (1).
Step 1: (28 \times 9=252) and (28 \times 10=280). Step 2: Since (280) is greater, the remainder is (275-252=23). Step 3: If the next multiple is greater, choose the previous multiple.
Step 1: (28 \times 9=252) and (28 \times 10=280). Step 2: Since (280) is greater than (275), the quotient is (9). Step 3: Choose the quotient whose product does not exceed the dividend.
Step 1: Adding (14) to the old remainder (12) gives (26). Step 2: (26=13 \times 2+0), so the new remainder is (0). Step 3: Do not forget to divide the new sum by the same divisor.
Step 1: (a-3=13q-2), but the remainder should not be negative. Step 2: (13q-2=13(q-1)+11), so the remainder is (11). Step 3: When a negative remainder appears, add the divisor to make the correct remainder.
Step 1: In (a=42q+41), the remainder is (41), one less than (42). Step 2: (a+1=42q+42=42(q+1)+0), so the remainder is (0). Step 3: Adding (1) to a (b-1) remainder gives exact division.
Step 1: In (a=42q+40), the remainder is (40). Step 2: (a+5=42q+45=42(q+1)+3), so the remainder is (3). Step 3: If the sum crosses the divisor, subtract the divisor once.
Step 1: On division by (31), the remainder must be from (0) to (30). Step 2: In (31q+30), the remainder is (30), which is less than (31). Step 3: A remainder (31), (37), or negative is not in standard form.
QUIZ COMPLETE
