Fundamental Theorem of Arithmetic

Step 1: For a perfect cube, every exponent must be a multiple of 3. Step 2: We must make (2^7), (3^5), and (5^4) into powers 9, 6, and 6. Step 3: So the smallest multiplier is (2^2\times3\times5^2).

Step 1: In multiplication, exponents of the same prime base are added. Step 2: The power of 3 in (a) is 4 and in (b) is 6. Step 3: In (ab), the power of 3 will be (4+6=10).

Step 1: In (xy), powers with the same base 5 are added. Step 2: The power of 5 in (x) is 5 and in (y) is 2. Step 3: The total power is (5+2=7).

Step 1: HCF uses only common prime factors. Step 2: The common factors are 2 and 3, with smaller powers (2^7) and (3^6). Step 3: Therefore, the HCF is (2^7\times3^6).

Step 1: LCM uses the highest powers of all prime factors. Step 2: The highest powers are (2^{10}), (3^8), (5^2), and (7^3). Step 3: So the correct form is (2^{10}\times3^8\times5^2\times7^3).

Step 1: Product of the two numbers is (48\times3360=161280). Step 2: One number is 240, so the other is (161280\div240=672). Step 3: As a check, the HCF of 240 and 672 is 48.

Step 1: Product of the two numbers is (72\times2520=181440). Step 2: The other number is (181440\div360=504). Step 3: To check, the HCF of 360 and 504 is 72.

Step 1: Prime factorise 7560. Step 2: (7560=756\times10=2^3\times3^3\times5\times7). Step 3: Comparing with the given form gives (a=3).

Step 1: Write (15120=16\times945). Step 2: (945=3^3\times5\times7), so (15120=2^4\times3^3\times5\times7). Step 3: Comparing gives (b=3).

Step 1: To count with repetition, add the exponents. Step 2: (2^7) gives 7, (3^5) gives 5, (5^4) gives 4, and 11 gives 1 factor. Step 3: Total (7+5+4+1=17), so the answer is 17.

Step 1: When counting distinct prime factors, do not add exponents. Step 2: The prime factors here are 2, 3, 5, and 11. Step 3: Therefore, the number of distinct prime factors is 4.

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: For two numbers, product (=) HCF (\times) LCM. Step 3: Therefore, the LCM will be 2431.

Step 1: Divide 16384 repeatedly by 2. Step 2: (16384=2^{14}). Step 3: 4 and 16 are composite, so they should not be written in the final prime form.

Step 1: Recognise (44100=210^2). Step 2: Since (210=2\times3\times5\times7), (210^2=2^2\times3^2\times5^2\times7^2). Step 3: In a perfect square, every prime exponent is even.

Step 1: (5265=3^4\times5\times13). Step 2: All these factors are present in the prime factorisation of (n). Step 3: Therefore, (n) must be divisible by 5265.

Step 1: In (n), the powers are (2^6), (3^2), and (5^2). Step 2: (675=3^3\times5^2), which needs power 3 of 3. Step 3: Since (n) has only (3^2), (n) will not be divisible by 675.

Step 1: Calculate (2^8=256) and (3^3=27). Step 2: (256\times27\times5=34560). Step 3: In larger products, simplifying powers first is safer.

Step 1: Calculate (2^4=16), (3^5=243), and (5^3=125). Step 2: (16\times243\times125=486000). Step 3: Simplifying powers first makes the calculation easier.

Step 1: In LCM, take the higher power of each prime. Step 2: The powers of 2 are 4 and 7. Step 3: The higher power is 7, so the answer is 7.