Fundamental Theorem of Arithmetic

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.

Step 1: Recognise (11025=105^2). Step 2: Since (105=3\times5\times7), (105^2=3^2\times5^2\times7^2). Step 3: In perfect squares, every prime exponent is even.

Step 1: The prime factorisation of 315 is (3^2\times5\times7). Step 2: All these factors are present in the prime factorisation of (n). Step 3: Therefore, (n) must be divisible by 315.

Step 1: In (n), the powers are (2^4), (3^1), and (5^2). Step 2: (48=2^4\times3), so divisibility by 48 is possible. Step 3: (80=2^4\times5), (75=3\times5^2), and (150=2\times3\times5^2) are also present; hence all given options divide (n).

Step 1: Calculate (2^6=64) and (3^2=9). Step 2: (64\times9\times5=2880). Step 3: In larger products, evaluate powers first and then multiply.

Step 1: Calculate (2^3=8), (3^4=81), and (5^2=25). Step 2: (8\times81\times25=16200). Step 3: Simplifying powers first keeps the calculation safe.

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