Fundamental Theorem of Arithmetic

Step 1: Prime factorise 180. Step 2: (180=18\times10=2^2\times3^2\times5). Step 3: Comparing with the given form gives (a=2).

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: The prime factors of the first number are 2 and 7. Step 2: The prime factors of the second number are 3 and 5. Step 3: There is no common prime factor, so they are co-prime.

Step 1: (28=2^2\times7) and (45=3^2\times5), so they are co-prime. Step 2: The LCM of co-prime numbers equals their product. Step 3: (28\times45=1260), so the answer is 1260.

Step 1: Divide 2048 repeatedly by 2. Step 2: (2048=2^{11}). Step 3: 4 and 32 are composite, so do not use them in the final prime form.

Step 1: Calculate (2^2=4) and (3^4=81). Step 2: (4\times81\times5=1620). Step 3: Simplifying powers first makes multiplication easier.

Step 1: Write (1125=45\times25). Step 2: (45=3^2\times5) and (25=5^2), so (1125=3^2\times5^3). Step 3: 45 and 25 are composite, so write their prime powers in the final form.

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

Step 1: For HCF, take smaller powers. Step 2: The smaller power of 2 is 2 and of 3 is 3. Step 3: (2^2\times3^3=4\times27=108), so the answer is 108.

Step 1: For LCM, take the highest powers. Step 2: The highest powers are (2^4) and (3^5). Step 3: (16\times243=3888), so the answer is 3888.

Step 1: Write (1365=105\times13). Step 2: (105=3\times5\times7), so (1365=3\times5\times7\times13). Step 3: The distinct prime factors are 3, 5, 7, and 13.

Step 1: For two numbers, product (=) HCF (\times) LCM. Step 2: (28\times420=11760). Step 3: In this relation, multiply the two given values directly.

Step 1: For two numbers, product (=) HCF (\times) LCM. Step 2: HCF (=5040\div360=14). Step 3: After finding the answer, you can check by multiplying.

Step 1: Write (3375=27\times125). Step 2: (27=3^3) and (125=5^3), so (3375=3^3\times5^3). Step 3: 27 and 125 are composite, so keep prime powers in the final form.

Step 1: A prime number must have exactly two positive factors. Step 2: 1 has only one positive factor, so it is not prime. Step 3: A composite number needs more than two factors, so 1 is not composite either.

Step 1: The common prime factors are 2 and 3. Step 2: The smaller powers are (2^2) and (3^1). Step 3: (2^2\times3=4\times3=12), so the HCF is 12.

Step 1: For LCM, take the highest powers. Step 2: The highest powers are (2^3), (3^2), (5), and (7^2). Step 3: (8\times9\times5\times49=17640), so the answer is 17640.

Step 1: Write (2772=252\times11). Step 2: (252=2^2\times3^2\times7), so (2772=2^2\times3^2\times7\times11). Step 3: 252 is composite, so it should not remain in the final form.

Step 1: The theorem is about prime factorisation of numbers greater than 1. Step 2: This factorisation is unique except for order. Step 3: Do not mix it with separate rules of HCF or co-primality.

Step 1: Calculate (2^5=32), (3^2=9), and (5^2=25). Step 2: (32\times9\times25=7200). Step 3: In larger multiplication, simplify powers first.