Muft Shiksha™ एक 100% Free Education Portal है 🇮🇳, जिसका उद्देश्य Class 9–12 के हर विद्यार्थी तक High-Quality Education को पूरी तरह मुफ्त पहुँचाना है। 🇮🇳 हम मानते हैं कि अच्छी शिक्षा किसी student की आर्थिक स्थिति पर निर्भर नहीं होनी चाहिए। 🇮🇳 हर विद्यार्थी को वही Quality Study Material, MCQs, Quizzes, Exam Preparation, Concept-Based Learning और Bilingual Support मिलना चाहिए, जो आमतौर पर महंगी Coaching या Premium Platforms में मिलता है। Muft Shiksha™ 🇮🇳 इसी सोच के साथ बनाया गया है
SubjectsMathematics
अंकगणित का मौलिक प्रमेय
In this Class 10 Mathematics topic from the chapter Real Numbers, students learn that every integer greater than 1 can be expressed as a product of prime numbers, and that this prime factorisation is unique apart from the order of the factors. They practise finding prime factors and use the theorem to understand and determine the HCF and LCM of numbers. The topic builds clear reasoning about the structure of whole numbers and supports later work with divisibility and number relationships.
TOPIC PRACTICE
Up to 25 questions from this page. Select your focus, then start.
Step 1: The Fundamental Theorem of Arithmetic states uniqueness of prime factorisation. Step 2: The order of prime factors may change, but the prime factors themselves do not change. Step 3: In exams, do not treat a change of order as a different factorisation.
Step 1: Write (27720=2520\times11). Step 2: Since (2520=2^3\times3^2\times5\times7), the full factorisation is (2^3\times3^2\times5\times7\times11). Step 3: Do not leave a composite factor like 2520 in the final answer.
Step 1: For HCF, take the smaller powers of only the common prime factors. Step 2: The common prime factors are 2 and 3, with smaller powers (2^2) and (3^3). Step 3: (2^2\times3^3=4\times27=108), so the answer is 108.
Step 1: For LCM, take the highest powers of all prime factors. Step 2: The highest powers are (2^4), (3^5), (5^2), and (7). Step 3: (16\times243\times25\times7=680400), so the answer is 680400.
Step 1: For two numbers, product (=) HCF (\times) LCM. Step 2: Therefore, LCM (=60480\div72=840). Step 3: Use this relation directly only for two numbers.
Step 1: For two numbers, product (=) HCF (\times) LCM. Step 2: (45\times1260=56700). Step 3: In such questions, first notice that exactly two numbers are involved.
Step 1: (1800=18\times100=2^3\times3^2\times5^2). Step 2: For a perfect square, all exponents must be even, but the exponent of 2 is 3. Step 3: Multiplying by 2 makes it 4, so the smallest number is 2.
Step 1: (8820=2^2\times3^2\times5\times7^2). Step 2: For a perfect square, all exponents must be even, but the exponent of 5 is 1. Step 3: Dividing by 5 makes all remaining exponents even, so the smallest number is 5.
Step 1: (5400=54\times100=2^3\times3^3\times5^2). Step 2: For a perfect cube, each prime exponent must be a multiple of 3. Step 3: The exponent of 5 is 2, so one more 5 is needed; the smallest number is 5.
Step 1: Recognise (9261=21^3). Step 2: Since (21=3\times7), (21^3=3^3\times7^3). Step 3: (21^3) gives the value, but 21 is not prime, so write the final prime form separately.
Step 1: For a perfect square, every exponent must be even. Step 2: The exponents of (2^5) and (7^1) are odd, while the others are even. Step 3: Multiplying by (2\times7=14) makes all exponents even.
Step 1: For a perfect cube, every exponent must be a multiple of 3. Step 2: To make powers (4,5,2) into (6,6,3), we need (2^2), (3), and (5). Step 3: So the smallest multiplier is (2^2\times3\times5).
Step 1: In multiplication, exponents of the same prime base are added. Step 2: The power of 3 in (a) is 2 and in (b) is 4. Step 3: In (ab), the power of 3 will be (2+4=6).
Step 1: In (xy), the powers of the same base 5 are added. Step 2: The power of 5 in (x) is 1 and in (y) is 4. Step 3: The total power is (1+4=5).
Step 1: HCF uses only common prime factors. Step 2: The common factors are 2 and 3; the smaller powers are (2^5) and (3^4). Step 3: Therefore, the HCF is (2^5\times3^4).
Step 1: LCM uses the highest powers of all prime factors. Step 2: The highest powers are (2^8), (3^6), (5^2), and (7). Step 3: So the correct form is (2^8\times3^6\times5^2\times7).
Step 1: For two numbers, product (=) HCF (\times) LCM. Step 2: Product of the two numbers is (18\times540=9720). Step 3: The other number is (9720\div90=108).
Step 1: Product of the two numbers is (24\times840=20160). Step 2: One number is 120, so the other is (20160\div120=168). Step 3: You can check the answer by confirming that HCF of 120 and 168 is 24.
Step 1: Prime factorise 720. Step 2: (720=72\times10=2^4\times3^2\times5). Step 3: Comparing with the given form gives (a=4).
Step 1: Write (1512=8\times189). Step 2: (8=2^3) and (189=3^3\times7), so (1512=2^3\times3^3\times7). Step 3: Comparing gives (b=3).
Step 1: To count with repetition, add the exponents. Step 2: (2^4) gives 4, (3^3) gives 3, and (5^2) gives 2 factors. Step 3: Total (4+3+2=9), so the answer is 9.
Step 1: When counting distinct prime factors, do not add exponents. Step 2: The prime factors here are 2, 3, and 7. Step 3: Therefore, the number of distinct prime factors is 3.
Step 1: The first number has prime factors 2 and 3. Step 2: The second number has prime factors 5 and 7. Step 3: There is no common prime factor, so they are co-prime.
Step 1: Co-prime numbers have HCF 1. Step 2: Product (=) HCF (\times) LCM. Step 3: Therefore, the LCM will be (1001).
Step 1: Divide 4096 repeatedly by 2. Step 2: (4096=2^{12}). Step 3: (4^6) and (16^3) can give the value, but 4 and 16 are not prime.
QUIZ COMPLETE
