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SubjectsMathematics
अभाज्य गुणनखंडन द्वारा महत्तम समापवर्तक और लघुत्तम समापवर्त्य
In this Class 10 Mathematics topic from Real Numbers, students learn to find the HCF and LCM of two or more numbers through prime factorisation. They break each number into prime factors, compare the resulting powers, and select the appropriate common factors: the lowest powers for HCF and the highest powers for LCM. The topic also develops accuracy in writing factor trees, simplifies comparison between methods, and helps students apply these concepts to numerical and everyday problem-solving situations.
TOPIC PRACTICE
Up to 25 questions from this page. Select your focus, then start.
Step 1: The highest power of (2) in the LCM must be (5). Step 2: The second number has power (3), so (a=5) gives the highest power (5). Step 3: For LCM, check the maximum-power condition.
Step 1: Product (=) HCF (\times) LCM. Step 2: LCM (=\frac{9504}{12}=792). Step 3: You can simplify (12) as (3\times4) while dividing.
Step 1: Product of two numbers equals HCF times LCM. Step 2: LCM (=\frac{140\times196}{28}=980). Step 3: Divide (196) by (28) first to get (7), then multiply (140\times7).
Step 1: HCF takes the smaller power of each common prime. Step 2: The smaller powers of (2), (3), and (7) are (3), (1), and (1), so HCF (=2^3\times3\times7). Step 3: Match each smaller power with the correct base.
Step 1: HCF contains only the common prime factors. Step 2: The common primes are (2) and (3), with smaller powers (2^5) and (3^2). Step 3: For HCF, always take the smaller powers.
Step 1: Prime factorise: (63=3^2\times7), (98=2\times7^2), and (154=2\times7\times11). Step 2: The highest powers are (2), (3^2), (7^2), and (11), so the LCM is (9702). Step 3: Include a prime even if it appears in only one number.
Step 1: For two numbers, product (=) HCF (\times) LCM. Step 2: LCM (=\frac{66528}{72}=924). Step 3: For large numbers, simplify the division in small steps.
Step 1: Product of two numbers equals HCF times LCM. Step 2: The other number is (\frac{48\times1440}{240}=288). Step 3: First simplify (1440) by (240) to make the calculation easier.
Step 1: The smallest number exactly divisible by all given numbers is their LCM. Step 2: (54=2\times3^3), (72=2^3\times3^2), and (90=2\times3^2\times5), so LCM (=2^3\times3^3\times5=1080). Step 3: Choose the highest powers carefully.
Step 1: The greatest common divisor is the HCF. Step 2: (225=3^2\times5^2), (375=3\times5^3), and (525=3\times5^2\times7), so HCF (=3\times5^2=75). Step 3: Use only the smallest powers common to all numbers.
Step 1: HCF is (2^2\times3^2\times13), and LCM is (2^4\times3^5\times5\times13). Step 2: On division, subtract the powers of the same bases, giving (2^2\times3^3\times5). Step 3: In prime-power division, subtract exponents.
Step 1: The next common signal time is the LCM of the intervals. Step 2: (16=2^4), (24=2^3\times3), and (40=2^3\times5), so LCM (=2^4\times3\times5=240). Step 3: For repeated-time questions, use LCM.
Step 1: HCF uses the smaller power of each common prime. Step 2: The smaller power of (2) must be (5), so (a=5) is possible; the smaller power of (3) must be (3), so (b=3) is possible. Step 3: Check unknown powers separately.
Step 1: LCM contains the highest power of each prime. Step 2: The highest power of (3) must be (4), so (a=4); the highest power of (7) must be (3), so (b=3). Step 3: For LCM, focus on the maximum-power condition.
Step 1: Subtracting (9) makes the number divisible by (28), (42), and (56). Step 2: (28=2^2\times7), (42=2\times3\times7), and (56=2^3\times7), so LCM (=2^3\times3\times7=168). Hence the number is (168+9=177). Step 3: Add the common remainder at the end.
Step 1: The maximum number of identical boxes is found using HCF. Step 2: (168=2^3\times3\times7) and (252=2^2\times3^2\times7), so HCF (=2^2\times3\times7=84). Step 3: For maximum equal distribution, use HCF.
Step 1: Coprime numbers have HCF (1). Step 2: Product (=) HCF (\times) LCM, so product (=1\times667=667). Step 3: For coprime numbers, the LCM equals the product.
Step 1: HCF (=2^3\times11=88). Step 2: LCM (=2^4\times3\times11=528), so the sum is (88+528=616). Step 3: When sum is asked, find both values separately.
Step 1: The smallest such number is the LCM of the two given numbers. Step 2: The powers of (3) are (2) and (5), so the higher power (5) will be used. Step 3: For divisibility, choose the higher power.
Step 1: After taking out HCF (18), (r) and (s) are coprime. Step 2: LCM (=18rs=990), so (rs=55). Step 3: Divide by the HCF to simplify such questions.
Step 1: (132=2^2\times3\times11) and (308=2^2\times7\times11). Step 2: The common smaller powers are (2^2) and (11), so HCF (=44). Step 3: Compare prime factors before choosing the statement.
Step 1: LCM takes the higher power of a common prime. Step 2: The powers of (2) are (6) and (4), so the higher power is (6). Step 3: In LCM, powers are not added; only the higher power is taken.
Step 1: Check prime factors: (66=2\times3\times11), (88=2^3\times11), and (121=11^2). Step 2: The distinct primes in the LCM are (2), (3), and (11), so the count is (3). Step 3: Do not count powers as separate primes.
Step 1: The HCF must exactly divide the LCM. Step 2: (1260) is not exactly divisible by (45), so such whole numbers are not possible. Step 3: Check this necessary condition before searching for pairs.
Step 1: For maximum equal length, find the HCF. Step 2: (80=2^4\times5) and (128=2^7), so HCF (=2^4=16). Step 3: For maximum equal cutting, use HCF.
QUIZ COMPLETE
