If (a=6q+1), what is the remainder when (a+4) is divided by (6)?
Step 1: In (a=6q+1), the old remainder is (1). Step 2: (a+4=6q+5), so the new remainder is (5). Step 3: If the new remainder remains less than the divisor, it is the answer.
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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: In (a=6q+1), the old remainder is (1). Step 2: (a+4=6q+5), so the new remainder is (5). Step 3: If the new remainder remains less than the divisor, it is the answer.
Step 1: Compare with (a=bq+r). Step 2: The number multiplied with (8) is (9), so (9) is the quotient. Step 3: To identify the quotient, look at the multiplier written with the divisor.
Step 1: In the Euclidean form, (b) is the divisor. Step 2: In (75=8 \times 9+3), (8) is the number by which division is done. Step 3: Identify the divisor by looking at the first number in the product.
Step 1: A number divided by itself gives quotient (1). Step 2: (28=28 \times 1+0), so (q=1) and (r=0). Step 3: When dividend and divisor are the same, remember that the remainder is (0).
Step 1: Remainder (1) on division by (2) means the number has the form (2q+1). Step 2: A number of the form (2q+1) is odd. Step 3: To identify an odd number, focus on remainder (1).
Step 1: Remainder (0) means the number is exactly divisible by (2). Step 2: A number exactly divisible by (2) is even. Step 3: The general form of an even number is (2q).
Step 1: (a=3q+2) is given. Step 2: (a+1=3q+3=3(q+1)), so it is exactly divisible by (3). Step 3: While changing the form, add the added (1) to the remainder.
Step 1: (40 \times 12=480) and (40 \times 13=520). Step 2: Since (520) is greater, the remainder is (518-480=38). Step 3: Avoid the next larger multiple and always use the nearest smaller multiple.
Step 1: (30) cannot be the remainder because it is greater than (29). Step 2: (29q+30=29(q+1)+1), so the correct remainder is (1). Step 3: Always keep the remainder between (0) and (b-1).
Step 1: Match (a=29q+28) with (a=bq+r). Step 2: Here (r=28), and (28<29), so it is a valid remainder. Step 3: If the remainder is less than the divisor, the form is already correct.
Step 1: The main form of Euclid’s division lemma is (a=bq+r). Step 2: The remainder is greater than or equal to (0) and less than the divisor. Step 3: In theory questions, remembering the range of the remainder is important.
Step 1: (0 \times 72+45=45), so the dividend is (45). Step 2: The divisor is (72), and (45<72), so the quotient is (0) and the remainder is (45). Step 3: When the quotient is (0), the dividend is usually smaller than the divisor.
Step 1: In (a=20q+19), the remainder is (19). Step 2: (a+1=20q+20=20(q+1)+0), so the remainder is (0). Step 3: Adding (1) to a remainder (b-1) makes the new remainder (0).
Step 1: (a=20q+18). Step 2: (a+5=20q+23=20(q+1)+3), so the remainder is (3). Step 3: If the sum exceeds (20), subtract (20) to get the new remainder.
Step 1: (37 \times 27=999). Step 2: So the quotient is (27) and the remainder is (0). Step 3: If the product is exactly equal to the dividend, that multiplier is the quotient and the remainder is (0).
Step 1: (37 \times 27=999). Step 2: (999-999=0), so the remainder is (0). Step 3: In exact division, the remainder is always (0).
Step 1: The Euclidean form is (a=bq+r). Step 2: Here the divisor is (13) and the remainder is (8), so the form is (13q+8). Step 3: In a general form, place the divisor with (q) and the remainder at the end.
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 correctly as (3(q+1)).
Step 1: Write the number as (4q+3). Step 2: The next number is (4q+4=4(q+1)+0). Step 3: After the greatest remainder, the next number has remainder (0) again.
Step 1: The number has the form (5q). Step 2: The next number is (5q+1), so the remainder is (1). Step 3: In consecutive numbers, remainders increase in order and then return to (0).
Step 1: (9 \times 9=81) and (9 \times 10=90). Step 2: Since (90) is greater than (85), the quotient is (9). Step 3: For the quotient, take the greatest multiplier whose product does not exceed the dividend.
Step 1: Taking (q=9), we get (9 \times 9=81). Step 2: (85-81=4), so (r=4). Step 3: After finding the remainder, check that (4<9).
Step 1: When (r=0), the equation becomes (a=bq). Step 2: This means (a) is a multiple of (b). Step 3: Remember zero remainder as a sign of exact divisibility.
Step 1: (50 \times 13=650). Step 2: (682-650=32), so the remainder is (32). Step 3: In division by (50), the remainder must be less than (50).
Step 1: In (a=50q+49), the remainder is (49). Step 2: (a+2=50q+51=50(q+1)+1), so the remainder is (1). Step 3: If the new remainder exceeds the divisor, subtract the divisor from it.
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