If a number leaves remainder (6) when divided by (7), what happens after adding (1)?
Step 1: The original number is (7q+6). Step 2: Adding (1) gives (7q+7=7(q+1)). Step 3: Therefore, the new number is exactly divisible by (7).
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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: The original number is (7q+6). Step 2: Adding (1) gives (7q+7=7(q+1)). Step 3: Therefore, the new number is exactly divisible by (7).
Step 1: The number is of the form (9q+8). Step 2: Adding (1) gives (9q+9=9(q+1)). Step 3: Therefore, the new remainder is (0).
Step 1: (4 \times 8=32) and (4 \times 9=36). Step 2: (36) is greater than (35), so (q=8). Step 3: (35-32=3), so (r=3).
Step 1: The lemma gives the remainder range (0 \le r < b). Step 2: This means the remainder is always less than the divisor. Step 3: In statement-based questions, this rule is very useful.
Step 1: (r=0) means nothing is left after division. Step 2: Then (a=bq), so (a) is exactly divisible by (b). Step 3: Use zero remainder to identify divisibility.
Step 1: (14 \times 7=98). Step 2: (99-98=1), so the remainder is (1). Step 3: The nearest smaller multiple helps find the remainder quickly.
Step 1: (14 \times 7=98) and (14 \times 8=112). Step 2: (112) is greater than (99), so the quotient is (7). Step 3: For the quotient, choose a multiple less than or equal to the dividend.
Step 1: For remainder (5), the number should be of the form (12q+5). Step 2: (29=12 \times 2+5), so its remainder is (5). Step 3: Check each option by division.
Step 1: To leave remainder (3) on division by (7), the form should be (7q+3). Step 2: (45=7 \times 6+3). Step 3: The remainder is less than the divisor, so (45) is correct.
Step 1: A number exactly divisible by (10) is of the form (10q). Step 2: Such a number has units digit (0). Step 3: In division by (10), use the units digit for a quick decision.
Step 1: Compare with (a=bq+r). Step 2: In (84=13 \times 6+6), (q=6) and (r=6). Step 3: Remainder (6) is less than divisor (13), so the form is correct.
Step 1: On division by (16), remainders can be from (0) to (15). Step 2: These are (16) values in total. Step 3: For a divisor (b), there are (b) possible remainders.
Step 1: Putting (q=0) gives (a=b \times 0+r), so (a=r). Step 2: Since (r<b) is necessary, (a<b). Step 3: When the dividend is smaller than the divisor, the quotient can be (0).
Step 1: The dividend (5) is smaller than the divisor (9). Step 2: So the quotient is (0) and the remainder is (5). Step 3: Note that remainder (5) is less than (9).
Step 1: The original number is (3q+2). Step 2: Adding (1) gives (3q+3=3(q+1)). Step 3: Therefore, the new number is exactly divisible by (3).
Step 1: (10 \times 12=120). Step 2: (121-120=1), so the remainder is (1). Step 3: In division by (10), the units digit helps find the remainder.
Step 1: (10 \times 12=120) and (10 \times 13=130). Step 2: (130) is greater than (121), so the quotient is (12). Step 3: For the quotient, always take the nearest smaller multiple.
Step 1: The correct remainder must be at least (0) and less than (9). Step 2: In (70=9 \times 7+7), the remainder is (7), which is less than (9). Step 3: Negative or too large remainders are not accepted.
Step 1: (6 \times 10=60) and (6 \times 11=66). Step 2: (60) is the correct smaller multiple and (64-60=4). Step 3: Since (4<6), (64=6 \times 10+4) is the correct form.
Step 1: The number is of the form (11q+10). Step 2: Adding (1) gives (11q+11=11(q+1)). Step 3: Therefore, the new remainder is (0).
Step 1: The remainder range is (0 \le r < b). Step 2: Here (r=21) and (b=21), so the condition (r<b) is not satisfied. Step 3: A remainder is never written equal to the divisor.
Step 1: Compare with the form (a=bq+r). Step 2: (16) is the divisor, (9) is the quotient, and (6) is the remainder. Step 3: Remainder (6) is less than (16), so the form is correct.
Step 1: The original number is (5q+1). Step 2: Adding (4) gives (5q+5=5(q+1)). Step 3: The new number is exactly divisible by (5), so the remainder is (0).
Step 1: (12 \times 6=72) and (12 \times 7=84). Step 2: (72) is the correct smaller multiple and (73-72=1). Step 3: Remainder (1) is less than (12), so the form is correct.
Step 1: The remainder condition is (0 \le r < b). Step 2: When (b=1), we get (0 \le r < 1), so only (r=0) is possible. Step 3: Any integer divided by (1) leaves remainder (0).
QUIZ COMPLETE