Which is the prime factorisation of 176400?
Step 1: Recognise (176400=420^2). Step 2: Since (420=2^2\times3\times5\times7), (420^2=2^4\times3^2\times5^2\times7^2). Step 3: In a perfect square, every prime exponent is even.
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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: Recognise (176400=420^2). Step 2: Since (420=2^2\times3\times5\times7), (420^2=2^4\times3^2\times5^2\times7^2). Step 3: In a perfect square, every prime exponent is even.
Step 1: (20655=3^5\times5\times17). Step 2: All these prime factors are present in (n) with sufficient powers. Step 3: Therefore, (n) must be divisible by 20655.
Step 1: In (n), the powers are (2^8), (3^4), and (5^2). Step 2: (3375=3^3\times5^3), which needs power 3 of 5. Step 3: Since (n) has only (5^2), (n) is not divisible by 3375.
Step 1: Calculate (2^9=512) and (3^4=81). Step 2: (512\times81\times5=207360). Step 3: In larger products, simplifying powers first is the right method.
Step 1: Calculate (2^5=32), (3^6=729), and (5^3=125). Step 2: (32\times729\times125=2916000). Step 3: Simplifying powers first keeps the calculation manageable.
Step 1: In LCM, take the higher power of each prime. Step 2: The powers of 2 are 5 and 8. Step 3: The higher power is 8, so the answer is 8.
Step 1: In HCF, take the smaller power of the common prime. Step 2: The powers of 3 are 6 and 3. Step 3: The smaller power is 3, so the answer is 3.
Step 1: Write (131220=26244\times5). Step 2: (26244=4\times6561=2^2\times3^8). Step 3: Therefore, (131220=2^2\times3^8\times5).
Step 1: For a perfect cube, exponents must be multiples of 3. Step 2: Make (2^{10}), (3^5), and (5^4) into powers 12, 6, and 6. Step 3: The smallest multiplier is (2^2\times3\times5^2).
Step 1: (63504=16\times3969=2^4\times3^4\times7^2). Step 2: All exponents are even, so the number is already a perfect square. Step 3: The smallest divisor should be 1, but 1 is not listed; therefore no given option is correct.
Step 1: In multiplication, exponents of the same prime base are added. Step 2: The power of 7 in (x) is 4 and in (y) is 5. Step 3: In (xy), the power of 7 is (4+5=9).
Step 1: Powers of the same base 2 are added in multiplication. Step 2: The power of 2 in (x) is 8 and in (y) is 7. Step 3: The total power is (8+7=15).
Step 1: Write (510510=30030\times17). Step 2: (30030=2\times3\times5\times7\times11\times13). Step 3: Therefore, the distinct prime factors are 2, 3, 5, 7, 11, 13, and 17.
Step 1: The common prime factors are 2, 3, and 11. Step 2: The smaller powers are (2^4), (3^3), and (11^1). Step 3: (16\times27\times11=4752), so the HCF is 4752.
Step 1: For LCM, take the highest powers of all prime factors. Step 2: The highest powers are (2^6), (3^5), (5^2), and (11^2). Step 3: (64\times243\times25\times121=47044800), so the correct value is 47044800.
Step 1: The product is (180\times25200). Step 2: (180=2^2\times3^2\times5) and (25200=2^4\times3^2\times5^2\times7). Step 3: The power of 2 in the product is (2+4=6).
Step 1: The product is (78\times510510). Step 2: (78=2\times3\times13), and 510510 also has 13 to power 1. Step 3: Therefore, the power of 13 in the product is (1+1=2).
Step 1: (83521=289\times289). Step 2: Since (289=17^2), (83521=17^4). Step 3: 289 is composite, so (289^2) is not the final prime factorisation.
Step 1: In a perfect cube, every exponent must be a multiple of 3. Step 2: Reduce (2^{10}) to (2^9) by dividing by 2, and (3^8) to (3^6) by dividing by (3^2). Step 3: The smallest divisor is (2\times3^2).
Step 1: First calculate (2^6=64) and (3^4=81). Step 2: (64\times81\times5\times7=181440). Step 3: To get the number from prime factorisation, multiply all factors.
Step 1: For two numbers, HCF (\times) LCM equals the product of the two numbers. Step 2: In prime powers, the smaller and higher exponents together give the total exponent. Step 3: Therefore, the answer is (ab).
Step 1: HCF is (2^6\times3^2). Step 2: LCM is (2^9\times3^5). Step 3: The ratio is (2^{9-6}\times3^{5-2}=2^3\times3^3=216).
Step 1: In (xy), exponents of the same bases are added. Step 2: Powers become (2^{12}), (3^9), and (5^9). Step 3: Counting with repetition gives (12+9+9=30).
Step 1: In LCM, take the higher power. Step 2: The powers of 5 are 3 and 6. Step 3: The higher power is 6, so the answer is 6.
Step 1: In HCF, take the smaller power. Step 2: The powers of 3 are 4 and 7. Step 3: The smaller power is 4, so the answer is 4.
QUIZ COMPLETE